GRADE

8

Math

Students are expected to build on their knowledge from the previous year, at a higher level of complexity. The depth of concepts, mathematical procedures and processes will be expanded each year. More homework, harder tests, and tough reading assignments—middle school and junior high is a big transition for your child. Tutoring is developed on an individualized plan that builds the academic skills, good habits and positive attitudes your child needs to succeed in middle school and beyond. Prerequisite: Math 7

Number Sense, Numeration, Rates, Ratios

Measurement

Angles, Coordinates, Geometry

Patterning and Algebra

Data Management and Probability

Number Sense, Numeration, Rates, Ratios

Representations of Powers: Pre-Product Rule

Simplify leaving your answer in the most reduced exponent form.

2 × 2 × 2 Solution

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= 2³

3 × 3 × 3 × 3 × 3 Solution

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= 3⁵

b × b × b Solution

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= b³

c × c × c × c × c Solution

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= c⁵

Scientific Notation: Expanded Form with Powers of × 10ⁿ

Convert to scientific notation.

148000 Solution

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× 10

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0.00000512 Solution

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× 10

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Understanding Powers of Ten with Scientific Notation

Convert the following scientific notation numbers.

1.725 × 10^-9

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(1.2 × 10²) + (3.4 × 10³)

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Comparing Fractions

Compare the following by filling in the blank (without using a calculator).

$\quad \dfrac{3}{6} \ \text{} \ \dfrac{2}{4}$ Solution

>

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None of the above

Make a common denominator in order to compare. Use the lowest common multiple (LCM) of 6 & 4: 12... $\begin{align} \dfrac{3}{6} & \text{} \dfrac{2}{4} \\ \\ \dfrac{3 × 2}{6 × 2} & \text{} \dfrac{2 × 3}{4 × 3} \\ \\ \dfrac{6}{12} & = \dfrac{6}{12} \\ \\ \end{align}$

$\quad \dfrac{16}{25} \ \text{____} \ \dfrac{7}{10}$ Solution Video

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None of the above

$\dfrac{16}{25} \ < \ \dfrac{7}{10}$

$\quad \dfrac{2}{3} \ \text{____} \ \dfrac{4}{5}$ Solution Video

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None of the above

Make a common denominator (with the lowest common multiple, LCM)
Multiples of 3: 3, 6, 9, 12, 15, ...
Multiples of 5: 5, 10, 15, ...
Therefore $\dfrac{2}{3} < \dfrac{4}{5}$

Converting Decimals and Fractions

Convert 0.75 to a fraction (remember to reduce fully). Solution

$\dfrac{7}{5}$

$\dfrac{3}{4}$

$\dfrac{75}{10}$

$\dfrac{3}{2}$

$\dfrac{3}{4}$

Converting Decimals and Fractions

Convert $\dfrac{250}{100}$ to a decimal. Solution

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2.5

Factors and Multiples

The following numbers are factors of 22 Solution Video 22, 44, 66, 88, 110

True

False

These are multiples of 22. Multiples multiply, going up higher each time... There are infinite amount of multiples...

(There are only a certain amount of factors... for 22 it's: 1, 2, 11, 22)

Determine the missing factor of 36: Solution 1, 2, 3, 4, 6, 9, 12, 18

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36 is a factor of 36 because you can divide 36 by 36.

Fraction and Percent Word Problems

Jon wins $100. He gives 50% to his mother, and from what is left over after giving money to his mother, he gives $\dfrac{2}{5}$ to charity. How much does Jon give to charity? Solution

$

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20 [steps not shown here]

Converting Decimals, Fractions, and Percents

Fill in the blank.

Solution

Fraction Percent

$\dfrac{2}{4}$ 50 %

$\dfrac{3}{4}$ 75 %

$\dfrac{2}{5}$

$\dfrac{1}{10}$ 10%

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%

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Fraction Percent

$\dfrac{2}{4}$ 50 %

$\dfrac{3}{4}$ 75 %

$\dfrac{2}{5}$ 40 %

$\dfrac{1}{10}$ 10%

Solution

Decimal Percent

0.3 30%

0.25 25%

5%

0.01 1%

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Decimal Percent

0.3 30%

0.25 25%

0.05 5%

0.01 1%

Word Problems with Percents with One Decimal or Fewer

Determine the price or percent below, using a calculator.

13.5% of $49.99 Solution

$

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$\begin{align} final\ amount & = \dfrac{percent}{100} \times \ initial\ amount \\ \\ & = \dfrac{13.5\%}{100\%} \times \ $49.99 \\ \\ & = 0.135 \times \ $49.99 \\ \\ & = $6.75 \end{align}$

125% of __ equals $50 Solution

$

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$\begin{align} final\ amount & = \dfrac{percent}{100} \times \ initial\ amount \\ \\ $50 & = \dfrac{125\%}{100\%} \times \ initial\ amount \\ \\ $50 & = 1.25 \times \ initial\ amount \end{align}$ If you are a bit advanced by now, and know algebra, then you can solve simply, $\begin{align} $50 & = 1.25 \times \ initial\ amount \\ \\ \dfrac{$50}{1.25} & = initial\ amount \\ \\ initial\ amount & = \dfrac{$50}{1.25} \\ \\ & = $40 \end{align}$ Otherwise use the slower, guess and check method (continually trying different numbers using estimation)... $\begin{align} $50 & = \dfrac{125\%}{100\%} \times \ initial\ amount \\ \\ $50 & = 1.25 \times \ \text{_} \\ \\ $50 & = 1.25 \times \ $40 \end{align}$

___ % of $200,000 equals $25,000 Solution

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%

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$\begin{align} final\ amount & = \dfrac{percent}{100} \times \ initial\ amount \\ \\ $25,000 & = \dfrac{percent}{100} \times $200,000 \\ \\ \end{align}$ If you are a bit advanced by now, and know algebra, then you can solve simply, $\begin{align} $25,000 & = \dfrac{percent}{100} \times $200,000 \\ \\ \dfrac{$25,000}{$200,000} & = \dfrac{percent}{100} \\ \\ \dfrac{$25,000}{$200,000}(100) & = percent \\ \\ (0.125)(100) & = percent \\ \\ percent & = 12.5\% \\ \\ \end{align}$ Otherwise use the slower, guess and check method (continually trying different numbers using estimation)... $\begin{align} $25,000 & = \dfrac{?}{100} \times $200,000 \\ \\ & = \ ... \\ \\ $25,000 & = \dfrac{12.5}{100} \times $200,000 \end{align}$

Word Problems with Percents and Rates

The equation for earnings (from hourly rates and commissions) is given below. Show your work (in your own notes) and use a calculator to solve.

$earnings \ = \ (time,\ hr)(hourly\ rate) \ + \ \dfrac{commission \ \%}{100}(sale,\ $)$

Someone works for 20 hours at $14/hr, and earns a 3% commission on $6,000 sales. Determine their earnings. Solution

$

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$\begin{align} earnings \ & = \ (time,\ hr)(hourly\ rate) \ + \ \dfrac{commission \ \%}{100}(sale,\ $) \\ \\ & = \ (20)(14) \ + \ \dfrac{3}{100}(6,000) \\ \\ & = $460 \end{align}$

An employee works 155 hours in a month and earns @4,085. If their hourly rate is $12/hr, and they earn 1.25% commission, determine their total sales, 'x' for that month. Solution

$

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$\begin{align} earnings \ & = \ (time,\ hr)(hourly\ rate) \ + \ \dfrac{commission \ \%}{100}(sale,\ $) \\ \\ 2,085 & = \ (155)(12) \ + \ \dfrac{1.25}{100}(x) \\ \\ 2,085 & = \ 1,860 \ + \ (0.0125)(x) \\ \\ 0.0125x & = 225 \\ \\ x & = \dfrac{225}{0.0125} \\ \\ & = $18,000 \end{align}$

In a work week, a young student, Jeremy works 42 hours at $9/hour. He earns $1,337.20 that week, on sales of $8720. Determine his percent commission. Solution

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%

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$\begin{align} earnings \ & = \ (time,\ hr)(hourly\ rate) \ + \ \dfrac{commission \ \%}{100}(sale,\ $) \\ \\ 1,337.2 & = \ (42)(9) \ + \ \dfrac{c \ \%}{100}(8,720) \\ \\ 1,337.2 & = \ (378) \ + \ 87.2c \\ \\ 87.2c & = 1,337.2 - 378 \\ \\ 87.2c & = 959.2 \\ \\ c & = \dfrac{959.2}{87.2} \\ \\ & = 11\ \% \end{align}$

A young student, Sydney, works 2 jobs, at job 'A' she earns $8/hour and a 3.2% commission on $6,000 sales. At job 'B' she earns $12/hr and 0% commission on $8,000 sales. If she works at job 'B' for 12 hours less than double her hours worked at job 'A', and her total earnings are $2,000, then determine her hours worked at job 'A'. Solution

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hours

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Make algebra expressions for her hours worked at job 'A' and at job 'B'. Let (x) be her hours worked at job 'A'
Let (2x - 12) be her hours worked at job 'B' Then extend the earnings equation to account for both jobs. [This could also be done with separate equations (not shown).] $\begin{align} earnings \ & = \ (time,\ hr)(hourly\ rate) \ + \ \dfrac{commission \ \%}{100}(sale,\ $) \ + \ (time,\ hr)(hourly\ rate) \\ \\ 2,000 & = \ (x)(8) \ + \ \dfrac{3.2}{100}(6,000) \ + \ (2x - 12)(12) \\ \\ 2,000 & = \ 8x \ + \ (0.032)(6,000) \ + \ (24x - 144) \\ \\ 2,000 & = 32x + 48 \\ \\ 32x & = 1,952 \\ \\ x & = \dfrac{1,952}{32} \\ \\ & = 61\ hours \end{align}$

Adding, Subtracting, Multiplying, and Dividing Fractions

Simplify the following fraction expressions without the use of a calculator. Reduce your answer fully.

$\dfrac{1}{4} + \dfrac{3}{5}$ Solution

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$\dfrac{\quad\quad}{\quad\quad}$

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Make the denominator with the lowest common multiple (LCM)...
4 → 8 → 12 → 16 → $\bbox[PowderBlue, 3pt]{20}$
5 → 10 → 15 → $\bbox[PowderBlue, 3pt]{20}$ $\begin{align} & = \dfrac{1}{4} + \dfrac{3}{5} \\ \\ & = \dfrac{1}{4} × 1 \ + \ \dfrac{3}{5} × 1 \\ \\ & = \dfrac{1}{4} × \bbox[Bisque, 3pt]{\dfrac{5}{5}} \ + \ \dfrac{3}{5} × \bbox[Bisque, 3pt]{\dfrac{4}{4}} \\ \\ & = \dfrac{1 × 5}{4 × 5} + \dfrac{3 × 4}{5 × 4} \\ \\ & = \dfrac{5}{\bbox[PowderBlue, 3pt]{20}} + \dfrac{12}{\bbox[PowderBlue, 3pt]{20}} \\ \\ & = \dfrac{5 + 12}{20} \\ \\ & = \dfrac{17}{20} \\ \\ \end{align}$

$\dfrac{2}{3} + \dfrac{1}{2} - \dfrac{1}{4}$ Solution

Hint Clear Info

$\dfrac{\quad\quad}{\quad\quad}$

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Make the denominator with the lowest common multiple (LCM)...
3 → 6 → 9 → $\bbox[PowderBlue, 3pt]{12}$
2 → 4 → 6 → 8 → 10 → $\bbox[PowderBlue, 3pt]{12}$
4 → 8 → $\bbox[PowderBlue, 3pt]{12}$ $\begin{align} & = \dfrac{2}{3} + \dfrac{1}{2} - \dfrac{1}{4} \\ \\ & = \dfrac{2}{3} × \bbox[Bisque, 3pt]{\dfrac{4}{4}} + \dfrac{1}{2} × \bbox[Bisque, 3pt]{\dfrac{6}{6}} - \dfrac{1}{4} × \bbox[Bisque, 3pt]{\dfrac{3}{3}} \\ \\ & = \dfrac{2 × 4}{3 × 4} + \dfrac{1 × 6}{2 × 6} - \dfrac{1 × 3}{4 × 3} \\ \\ & = \dfrac{8}{12} + \dfrac{6}{12} - \dfrac{3}{12} \\ \\ & = \dfrac{8 + 6 - 3}{12} \\ \\ & = \dfrac{11}{12} \\ \\ \end{align}$

$2 × \dfrac{3}{4}$ Solution

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$\dfrac{\quad\quad}{\quad\quad}$

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You don't need a common denominator for multiplying fractions, look it's easy! $\begin{align} & = 2 × \dfrac{3}{4} \\ \\ & = \dfrac{2}{1} × \dfrac{3}{4} \\ \\ & = \dfrac{2 × 3}{1 × 4} \\ \\ & = \dfrac{6}{4} \\ \\ & = \dfrac{3}{2} \\ \\ \end{align}$

$\dfrac{1}{2} × \dfrac{2}{3} × \dfrac{1}{4}$ Solution

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$\dfrac{\quad\quad}{\quad\quad}$

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With multiplication of fractions you can cancel something once on the top and bottom... $\begin{align} & = \dfrac{1}{2} × \dfrac{2}{3} × \dfrac{1}{4} \\ \\ & = \dfrac{1 × 2 × 1}{2 × 3 × 4} \\ \\ & = \dfrac{1 × \cancel{2} × 1}{\cancel{2} × 3 × 4} \\ \\ & = \dfrac{1}{12} \\ \\ \end{align}$

$\dfrac{1}{2} ÷ \dfrac{1}{3}$ Solution

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$\dfrac{\quad\quad}{\quad\quad}$

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With division of fractions... multiply by the reciprocal of the fraction after the division sign ÷... $\begin{align} & = \dfrac{1}{2} ÷ \dfrac{1}{3} \\ \\ & = \dfrac{1}{2} × \dfrac{3}{1} \\ \\ & = \dfrac{1 × 3}{2 × 1} \\ \\ & = \dfrac{3}{2} \\ \\ \end{align}$

$\dfrac{3}{25} ÷ \dfrac{1}{5}$ Solution

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$\dfrac{\quad\quad}{\quad\quad}$

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With division of fractions... multiply by the reciprocal of the fraction after the division sign ÷ ... And cancel across division... $\begin{align} & = \dfrac{3}{25} ÷ \dfrac{1}{5} \\ \\ & = \dfrac{3}{25} × \dfrac{5}{1} \\ \\ & = \dfrac{3 × 5}{25 × 1} \\ \\ & = \dfrac{3 × \cancelto{1}{5}}{\cancelto{5}{25} × 1} \\ \\ & = \dfrac{3 × 1}{5 × 1} \\ \\ & = \dfrac{3}{5} \\ \\ \end{align}$

More Fractions with Order of Operations (PEMDAS/BEDMAS)

Simplify the following fraction expressions without the use of a calculator. Reduce your answer fully.

$\dfrac { 11 }{ 4 } \ - \ 1$ Solution

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$\dfrac{\quad\quad}{\quad\quad}$

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$\begin{align} & = \dfrac { 11 }{ 4 } \ - \ \color{RoyalBlue}{1} \\ \\ & = \dfrac { 11 }{ 4 } \ - \ \color{RoyalBlue}{\dfrac { 4 }{ 4 }} \\ \\ & = \dfrac { 11 \ - \ 4 }{ 4 } \\ \\ & = \dfrac { 7 }{ 4 } \\ \\ \end{align}$

$\left( \dfrac { 2 }{ 3 } +\dfrac { 1 }{ 2 } \right) \times \dfrac { 2 }{ 5 }$ Solution

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$\dfrac{\quad\quad}{\quad\quad}$

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$\begin{align} & = \left( \dfrac { 2 }{ 3 } + \dfrac { 1 }{ 2 } \right) \ \times \ \dfrac { 2 }{ 5 } \\ \\ & = \left( \dfrac { 4 }{ 6 } + \dfrac { 3 }{ 6 } \right) \ \times \ \dfrac { 2 }{ 5 } \\ \\ & = \left( \dfrac { 4 \ + \ 3 }{ 6 } \right) \ \times \ \dfrac { 2 }{ 5 } \\ \\ & = \dfrac { 7 }{ 6 } \ \times \ \dfrac { 2 }{ 5 } \\ \\ & = \dfrac { 7 \ \times \ 2 }{ 6 \ \times \ 5 } \\ \\ & = \dfrac { 14 }{ 30 } \\ \\ & = \dfrac { 7 }{ 15 } \\ \\ \end{align}$

$3 \ \div \ \dfrac { 6 }{ 11 }$ Solution

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$\dfrac{\quad\quad}{\quad\quad}$

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$\begin{align} & = 3 \ \color{WildStrawberry}{\div} \ \color{RoyalBlue}{\dfrac { 6 }{ 11 }} \\ \\ & = 3 \ \color{WildStrawberry}{\times} \ \color{RoyalBlue}{\dfrac { 11 }{ 6 }} \\ \\ & = \dfrac { 3 }{ 1 } \ \times \ \dfrac { 11 }{ 6 } \\ \\ & = \dfrac { 3 \ \times \ 11 }{ 1 \ \times \ 6 } \\ \\ & = \dfrac { 33 }{ 6 } \\ \\ & = \dfrac { 33 \ \color{Magenta}{\div 3} }{ 6 \ \color{Magenta}{\div 3} } \\ \\ & = \dfrac { 11 }{ 2 } \\ \\ \end{align}$

$\left( \dfrac { 2 }{ 9 } \ \div \ \dfrac { 4 }{ 3 } \right) \ \div \ \dfrac { 1 }{ 5 }$ Solution

Hint Clear Info

$\dfrac{\quad\quad}{\quad\quad}$

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$\begin{align} & = \left( \dfrac { 2 }{ 9 } \ \times \ \dfrac { 3 }{ 4 } \right) \ \div \ \dfrac { 1 }{ 5 } \\ \\ & = \left( \dfrac { 2 \ \times \ 3 }{ 9 \ \times \ 4 } \right) \ \div \ \dfrac { 1 }{ 5 } \\ \\ & = \left( \dfrac { 6 }{ 36 } \right) \ \div \ \dfrac { 1 }{ 5 } \\ \\ & = \left( \dfrac { 6 \ \color{ForestGreen}{\div \ 6} }{ 36 \ \color{ForestGreen}{\div \ 6} } \right) \ \div \ \dfrac { 1 }{ 5 } \\ \\ & = \left( \dfrac { 1 }{ 6 } \right) \ \color{Red}{\div} \ \dfrac { 1 }{ 5 } \\ \\ & = \dfrac { 1 }{ 6 } \ \color{Red}{\times} \ \dfrac { 5 }{ 1 } \\ \\ & = \dfrac { 1 \ \times \ 5 }{ 6 \ \times \ 1 } \\ \\ & = \dfrac { 5 }{ 6 } \\ \\ \end{align}$

Order of Operations (PEMDAS, BEDMAS)

Simplify the following, without the use of a calculator.

Solution 6 x (4 ÷ 2) + 8 + (10 + 4) – 2 × 6

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Use PEMDAS/BEDMAS...

Solution 3 × 4 ÷ 3 × 4 × (64 ÷ 16)

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Use PEMDAS/BEDMAS...

Solution Video (4 × 2) + 3² - 4 ÷ 2

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Use PEMDAS/BEDMAS... $\begin{align} & = (4 × 2) + 3^2 - 4 ÷ 2 \\ \\ & = (8) + 3^2 - 4 ÷ 2 \\ \\ & = 8 + 9 - 4 ÷ 2 \\ \\ & = 8 + 9 - 2 \\ \\ & = 15 \\ \\ \end{align}$

Multiply and Divide by Powers of Ten

Solve, without the use of a calculator.

25 × 10⁴ Solution

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Append 4 zeros to the end

You start with 25.0 and add 4 zeros to the end (move the decimal 4 places, right).

123456 ÷ 10³ =

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You start with 123456.0 and move the decimal 3 places, left, to make the number smaller.

200 ÷ 10^-5

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This is trick because the combination of '÷' with a negative exponents, is like a double negative, in that you end up moving the decimal place to the right. It's equivalent to 200 × 10⁵.

Squares and Roots

Simplify

$\sqrt{1^2}$ Solution

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A square root cancels out the 'perfect square' number. $\begin{align} & = \sqrt{1^2} \\ \\ & = 1 \\ \\ \end{align}$

$\sqrt{3^2}$ Solution

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$\begin{align} & = \sqrt{3^2} \\ \\ & = 3 \\ \\ \end{align}$

$\sqrt{4}\sqrt{81}$ Solution

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$\begin{align} & = (2)(9) \\ \\ & = 18 \\ \\ \end{align}$

$2\sqrt{5^2}\sqrt{15}$ Solution

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$\sqrt{\begin{matrix} \\ \quad \end{matrix}}$

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$\begin{align} & = 2\sqrt{5^2}\sqrt{15} \\ \\ & = 2(5)\sqrt{15} \\ \\ & = 10\sqrt{15} \\ \\ \end{align}$

Proportional Relationships

Solve the following real-life proportion problems. You can use a calculator, if needed.

A cookie recipe that makes 48 cookies calls for 2 eggs, 1 cup of butter, 2 1/4 cups of flour, and 2 cups of chocolate chips, and other stuff... Determine the integer ratio of flour to chocolate chips. Solution

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to

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$\begin{align} Ratio & = 2\ \dfrac{1}{4} : 2 \\ \\ & = \dfrac{9}{4} : 2 \\ \\ & = \dfrac{9}{4}(4) : 2(4) \\ \\ & = 9 : 8 \end{align}$

The density of typical, dry sand is given below. Determine the mass of 10 m³ of this sand. Solution $Density = 1,500\ \dfrac{kg}{m^3}$

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kg

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This is a directly proportional relationship, where the density is constant, so increasing one property (volume) by a certain amount will increase the other property (mass) by the same factor (× 10). You can determine this by guess-and-check, or some algebra... $\begin{align} Density & = \dfrac{mass\ kg}{volume\ m^3} \\ \\ 1500 & = \dfrac{mass\ kg × 10}{10\ m^3 × 10} \\ \\ & = \ ... \\ \\ mass & = 150,000\ kg \end{align}$

If the radius of a circular pancake is tripled, by what factor is the new circumference, and the new surface area increased? Solution Circumference = 2π·r

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Circumference:
Area:

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Circumference, $\begin{align} & = 2π(3) \end{align}$ Area, $\begin{align} & = π(3)^2 \\ \\ & = π(9) \end{align}$

Equivalent Fractions as Proportional Relationships

Solve, without reducing.

What fraction is to $\dfrac{3}{5}$ ? Solution

$\dfrac{11}{20}$

$\dfrac{18}{33}$

$\dfrac{35}{60}$

$\dfrac{33}{55}$

You can multiply the top (numerator) and bottom (denominator) by the same amount... 11 $\begin{align} & = \dfrac{3}{5} × 1 \\ \\ & = \dfrac{3}{5} × \dfrac{1}{1} \\ \\ & = \dfrac{3}{5} × \dfrac{11}{11} \\ \\ & = \dfrac{3 × 11}{5 × 11} \\ \\ & = \dfrac{33}{55} \\ \\ \end{align}$

Determine the answer.

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$\dfrac{3}{4} \ × \ \dfrac{2}{2} = \dfrac{}{\quad\quad}$

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Determine parts of each fraction.

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$\dfrac{5}{6} \ × \ \dfrac{3}{\quad\quad} = \dfrac{\quad\quad}{18}$

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Determine the first fraction.

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$\dfrac{1}{3} \ × \ \dfrac{}{\quad\quad} = \dfrac{7}{21}$

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

It's all or nothing now!

Hint Clear Info

$\dfrac{3}{4} \ = \ \dfrac{6}{\quad\quad} \ = \ \dfrac{\quad\quad}{12} \ = \ \dfrac{12}{\quad\quad} \ = \ \dfrac{30}{\quad\quad} \ = \ \dfrac{\quad\quad}{100}$

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Measurement

Conversions Between m² and cm²

Given the rectangle,

Calculate the area of the rectangle, in square centimeters. Solution

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cm²

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Area = Length × Width $\begin{align} & = 20 \ cm × 10 \ cm \\ \\ & = 200\ cm^2 \\ \\ \end{align}$

Convert the area of the rectangle from square centimeters into square meters. Solution

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m²

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Know that the place value difference between m² and cm² is 4 places...

Convert to meters... move the decimal 4 places left. $\begin{align} & = 200\ cm^2 \\ \\ & = 200.0\ cm^2 \\ \\ & = 00200.0\ cm^2 \\ \\ & = 0.02\ m^2 \\ \\ \end{align}$ Or, convert the lengths to meters first, and then calculate the area:

10 cm --> 0.1 m

20 cm --> 0.2 m

$\begin{align} Area & = (length)(width) \\ \\ & = (0.1\ m)(0.2\ m) \\ \\ & = 0.02\ m^2 \\ \\ \end{align}$

Conversions of Square and Cubic Units

Convert the following, given,

1 m² = 10,000 cm²
1 m³ = 1,000,000 cm³

2.5 m³ into cm³ Solution

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cm³

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$\begin{align} & = 2.5\ m^3 \dfrac{1,000,000\ cm^3}{1\ m^3} \\ \\ & = 2,500,000\ cm^3 \end{align}$

663 cm³ into m³ Solution

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m³

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$\begin{align} & = 663\ cm^3 \dfrac{1\ m^3}{1,000,000\ cm^3} \\ \\ & = 0.000663\ m^3 \end{align}$

The volume (capacity), in m³, of a rectangular prism with length 3 cm, width 8 cm, and heigh 11 cm. Solution

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m³

Hint Unavailable

$\begin{align} Volume & = (length)(width)(height) \\ \\ & = (3\ cm)(8\ cm)(11\ cm) \\ \\ & = 264\ cm^3 \end{align}$ Convert to m³ $\begin{align} & = 264\ cm^3 \dfrac{1\ m^3}{1,000,000\ cm^3} \\ \\ & = 0.000264\ m^3 \end{align}$

Perimeter and Area (Circles)

The area of a circle is calculated with the equation, Area = πr²

Which of the following can be used to calculate the amount of material required to make a circular mirror.

Radius

Circumference

Diameter

Any of the above

$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Determine the area of a circle with a diameter of 5 cm. Solution

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cm²

Hint Unavailable

Radius = Diameter ÷ 2 $\begin{align} Area & = πr^2 \\ \\ & = π(2.5)^2 \\ \\ & = 19.6\ cm^2 \end{align}$

Determine the area of a circle with a circumference of 11 cm. Solution

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cm²

Hint Unavailable

First, calculate the radius, this uses algebra, which some students may learn later $\begin{align} Circumference & = 2πr \\ \\ \dfrac{Circumference}{2π} & = r \\ \\ \dfrac{11}{2π} & = r \\ \\ r & = 1.75\ cm \end{align}$ Next, use this radius to calculate the area. $\begin{align} Area & = πr^2 \\ \\ & = π(1.75)^2 \\ \\ & = 9.6\ cm^2 \end{align}$

Area (Triangles)

A smaller triangle with a height of 5 cm is quadrupled in each of its dimensions (base, height, hypotenuse). If the area of the bigger triangle is 400 cm², determine the area of the smaller triangle. Solution

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cm²

Hint Unavailable

Given h₁ = 5 cm, then... $\begin{align} h_2 & = 4(h_1) \\ \\ h_2 & = 4(5\ cm) \\ \\ h_2 & = 20\ cm \end{align}$ With the area of the bigger triangle $\begin{align} Area_{bigger\ triangle} & = \dfrac{1}{2}(b_2)(h_2) \\ \\ 400 & = \dfrac{1}{2}(b_2)(20) \\ \\ 400 & = 10 \, b_2 \\ \\ b_2 & = \dfrac{400}{10} \\ \\ b_2 & = 40\ cm \end{align}$ Now, with b₂ = 40 cm, then... $\begin{align} b_2 & = 4(b_1) \\ \\ 40\ cm & = 4(b_1) \\ \\ b_1 & = \dfrac{40\ cm}{4} \\ \\ b_1 & = 10\ cm \end{align}$ Now with the area of the smaller triangle $\begin{align} Area_{smaller\ triangle} & = \dfrac{1}{2}(b_1)(h_1) \\ \\ & = \dfrac{1}{2}(10\ cm)(5\ cm) \\ \\ & = (5\ cm)(5\ cm) \\ \\ & = 25\ cm^2 \end{align}$

Area (Triangles)

Determine the area of the triangle 'xhb' given the area of triangle 'abc' is 69.65 cm², 'a' is 10 cm, 'b' is 14 cm, and 'c' is 18 cm. Solution

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cm²

See that triangle 'xhb' and triangle 'abc' share the same height 'h'.

With area of triangle 'abc' $\begin{align} Area & = \dfrac{1}{2}(base)(height) \\ \\ Area & = \dfrac{1}{2}(a)(h) \\ \\ 69.65 & = \dfrac{1}{2}(10)(h) \\ \\ 69.65 & = 5(h) \\ \\ h & = \dfrac{69.65}{5} \\ \\ h & = 13.93\ cm \end{align}$ Use Pythagorean theorem to find length 'x' $\begin{align} h^2 + x^2 & = b^2 \\ \\ (13.93)^2 + x^2 & = (14)^2 \\ \\ x^2 & = 196 - 194.045 \\ \\ x^2 & = 1.955 \\ \\ \sqrt{x^2} & = \sqrt{1.955} \\ \\ x & = 1.398\ cm \end{align}$ Area of triangle 'xhb' $\begin{align} Area & = \dfrac{1}{2}(base)(height) \\ \\ & = \dfrac{1}{2}(x)(h) \\ \\ & = \dfrac{1}{2}(1.398)(13.93) \\ \\ & = 9.74\ cm^2 \end{align}$

Circle Area

A circular, robotic vacuum cleaner cleans a room with dimensions, 2 m × 3 m. Since the robot is circular, assume that it cannot clean the four corners of the room. If the robot is 20 cm across, determine the area of the room that cannot be cleaned by the robot, and the percent of the room that can be cleaned [round to the nearest tenth decimal]. Solution

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cm² %

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The area missed is the area of a square, minus the area of one quarter of a circle. Also, make sure that your units 'agree' all in cm...

Make a sketch to see that the area that cannot be reached by the vacuum is the area of a square, minus the area of one quarter of a circle. Radius 'r' = 10 cm $\begin{align} Area_{missed} & = Area_{square} - \dfrac{1}{4}Area_{circle} \\ \\ & = (10)(10) - \dfrac{1}{4} \pi (10)^2 \\ \\ & = 100 - 25\pi \\ \\ & = 21.5\ cm^2 \end{align}$ There are four corners, so the total area that cannot be reached by the vacuum is, $\begin{align} & = 4(21.5\ cm^2) \\ \\ & = 86\ cm^2 \end{align}$ And the percent of the (200 cm × 300 cm) room that gets cleaned: $\begin{align} Percent & = \dfrac{Area_{cleaned}}{Area_{total}} \times 100\% \\ \\ & = \dfrac{Area_{total} - Area_{missed}}{Area_{total}} \times 100\% \\ \\ & = \dfrac{(200)(300) - 86}{(200)(300)} \times 100\% \\ \\ & = \dfrac{59,914}{60,000} \times 100\% \\ \\ & = 99.9\ \% \end{align}$

Circles: Compound Shapes with Additive and Subtractive Methods

Given the area of the square part is 225 cm², determine the length of the semicircle, indicated by the dotted line. Solution

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cm

Hint Unavailable

First find side length 's' of the square, given the area $\begin{align} Area_{square} & = (side\ length)^2 \\ \\ 225 & = s^2 \\ \\ \sqrt{225} & = \sqrt{s^2} \\ \\ s & = 15\ cm \end{align}$ Next, see that the side length 's' of the square is equal to the diameter 'd' of the semicircle. $\begin{align} Circumference\ of\ full\ circle & = \pi (d) \\ \\ & = \pi (15\ cm) \end{align}$ The perimeter of the semicircle, is half of this circle, $\begin{align} Circumference\ of\ half\ circle & = \dfrac{15\pi\ cm}{2} \\ \\ & = 23.6\ cm \end{align}$

If the diameter of the larger circle is double the diameter of the smaller circle, and the perimeter of the smaller circle is 5π cm, calculate the shaded area. Solution

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cm²

Hint Unavailable

Given the perimeter of the smaller circle, determine its diameter 'd' $\begin{align} Perimeter & = \pi (d) \\ \\ 5\pi & = \pi (d) \\ \\ d & = \dfrac{5\pi}{\pi} \\ \\ & = \dfrac{5\cancel{\pi}}{\cancel{\pi}} \\ \\ & = 5\ cm \end{align}$ Determine the radius 'r' of both the larger, and smaller circles, based on the diameter 'd' of the smaller circle,
$\begin{align} r_{\, bigger} & = 2(d_{\, smaller}) \\ \\ & = 2(5\ cm) \\ \\ & = 10\ cm \end{align}$ $\begin{align} r_{\, smaller} & = \dfrac{d_{\, smaller}}{2} \\ \\ & = \dfrac{5\ cm}{2} \\ \\ & = 2.5\ cm \end{align}$
Calculate the shaded area, $\begin{align} A_{\, shaded} & = A_{\, bigger} - A_{\, smaller} \\ \\ & = \pi (r_{\, bigger})^2 - \pi (r_{\, smaller})^2 \\ \\ & = \pi (10\ cm)^2 - \pi (2.5\ cm)^2 \\ \\ & = 100\pi - 6.25\pi \\ \\ & = 294.5\ cm^2 \end{align}$

Circles: Compound Shapes with Additive and Subtractive Methods

For the following figure, all circles are equal diameter, and the length of the dotted semicircle is 6π cm.

Determine the radius 'r' of the circles. Solution

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cm

Hint Unavailable

Use the given length 'L' of semicircle, to calculate the radius. A semicircle is half the circumference of the whole circle. $\begin{align} Circumference & = 2 \times (semicircle\ length) \\ \\ & = 2(L) \\ \\ & = 2(6\pi\ cm) \\ \\ & = 12\pi\ cm \end{align}$ Calculate the radius 'r' $\begin{align} Circumference & = 2\pi r \\ \\ 12\pi & = 2\pi r \\ \\ r & = \dfrac{12\pi}{2\pi} \\ \\ & = \dfrac{\cancelto{6}{12}\cancel{\pi}}{\cancelto{1}{2}\cancel{\pi}} \\ \\ & = 6\ cm \end{align}$

Determine the perimeter of the rectangle. Solution

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cm

Hint Unavailable

Previously, the radius 'r' was calculated, as 6 cm. The height of the rectangle is equal to the diameter of the circles, $\begin{align} Height & = 2(6\ cm) \\ \\ & = 12\ cm \end{align}$ The width of the rectangle is equal to two circle halves (radii), plus one whole diameter. $\begin{align} Width & = 2(radius) + 1(diameter) \\ \\ & = 2(6\ cm) + 1(12\ cm) \\ \\ & = 24\ cm \end{align}$ Calculate the perimeter, $\begin{align} Perimeter & = 2(height) + 2(width) \\ \\ & = 2(12\ cm) + 2(24\ cm) \\ \\ & = 72\ cm \end{align}$

Calculate the area of all parts of the circles contained within the rectangle, Solution

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cm²

Hint Unavailable

See that the two semicircles, put together, make a whole. Plus, the other whole circle, so, $\begin{align} Area_{total} & = 2 \times (Area_{circle}) \\ \\ & = 2 \times [\pi (r)^2] \\ \\ & = 2 \times [\pi (6)^2] \\ \\ & = 72\ pi \\ \\ & = 226.2\ cm^2 \end{align}$

Calculate the shaded area between, but not including the circles, within the rectangle. Solution

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cm²

Hint Unavailable

$\begin{align} Area_{shaded} & = Area_{rectangle} - Area_{circle\ parts} \\ \\ & = (h)(w) - 2 \times [\pi (r)^2] \\ \\ & = (12)(24) - 2 \times [\pi (6)^2] \\ \\ & = 288 - 72\ pi \\ \\ & = 61.8\ cm^2 \end{align}$

Area with Compound Shapes

Use compound shapes to solve the following.

Area of the kite. Solution

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units²

Hint Unavailable

Split the kite into compound shapes: 2, identical triangles Calculate the area Yes, there is also an equation for kite, $\left[\dfrac{1}{2}(diagonal_1)(diagonal_2)\right]$ but this was an experiment in breaking down into compound shapes.

Area of the trapezoid. Solution

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units²

Hint Unavailable

Split the trapezoid into compound shapes: rectangle + square Calculate the area Yes, there is also an equation for trapezoid, $\left[\dfrac{1}{2}(height)(top + bottom)\right]$ but this was an experiment in breaking down into compound shapes.

Perimeter and Area

A circle and rectangle have the same area [note: not drawn to scale, so the heights are not necessarily equal]. The radius of the circle is 3cm, and the width of the rectangle is 6cm.

Determine the length of the rectangle, round your answer to one decimal place. Solution

length =

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The statement says the circle and rectangle have the same area: $\begin{align} Area_{rectangle} & = Area_{circle} \\ \\ l × w & = \pi r^2 \\ \\ l × (6) & = \pi (3)^2 \\ \\ 6l & = 9\pi \\ \\ l & = \dfrac{9\pi}{6} \\ \\ l & = 4.7\ cm \\ \\ \end{align}$

Determine which shape has a larger perimeter, show your work. Solution

The perimeters are the same

Rectangle

Square

Not enough information given

$\begin{align} Perimeter_{rectangle} & = 2l + 2w \\ \\ & = 2(4.7) + 2(6) \\ \\ & = 21.4\ cm \\ \\ \\ \\ \\ \\ Perimeter_{circle} & = 2\pi r \\ \\ & = 2\pi (3) \\ \\ & = 18.8\ cm \\ \\ \end{align}$ Therefore the rectangle has a larger perimeter.

Perimeter and Area: Factors, Multiples

A pizza place sells 8" (diameter), 12", and 16" pizzas. How many times larger is the area of the 16" than the 8" pizza? Solution

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times larger

Hint Unavailable

Remember this is area... proportional to the square of the radius. Doubling the radius or diameter is always equivalent to quadrupling (4×) the area, which is the amount of pizza that you get.

For example, $\begin{align} A_{8} & = \pi (r)^2 \\ \\ & = \pi (4)^2 \\ \\ & = 16\pi \\ \\ \\ \\ A_{16} & = \pi (8)^2 \\ \\ & = 64\pi \\ \\ \end{align}$ Hopefully you can see that the area of the largest pizza is 4x the area of the smallest pizza, $\begin{align} & = \dfrac{\cancelto{4}{64} \cancel{\pi}}{\cancelto{1}{16}\cancel{\pi}} \\ \\ & = 4 \\ \\ \end{align}$

Cylinder Surface Area

The surface area of the shape below can be calculated with which of the following formulas? Solution

$\pi r^2h + 2\pi r$

$\dfrac{4}{3}\pi r^3 + h$

$\pi r^2 + \pi rl$

$2\pi r^2 + 2\pi rh$

$2\pi r^2 + 2\pi rh$ is the formula for surface area of a cylinder. Think of the sum of the compound shapes.

The surface area of a cylinder is 226.2 cm², and the radius is 4 cm. Determine the height, showing your work. Solution

h =

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cm

Hint Unavailable

$\begin{align} SA & = 2\pi r^2 + 2\pi rh \\ \\ 226.2 & = 2\pi(4)^2 + 2\pi(4)h \\ \\ 226.2 & = 2(3.14)(4)^2 + 2(3.14)(4)h \\ \\ 226.2 & = 6.28(16) + 6.28(4)h \\ \\ 226.2 & = 100.48 + 25.12h \\ \\ 25.12h & = 226.2 - 100.48 \\ \\ 25.12h & = 125.72 \\ \\ h & = \dfrac{125.72}{25.12} \\ \\ h & = 5\ cm \\ \\ \end{align}$

Cylinder Surface Area

Paint coverage for a certain rust-resistance paint is 23 m². Determine the length 'L' of the open-ended pipe that can be painted, if the radius 'r' is 5 cm, and you only paint (all the way around) the curved part of the pipe. Solution

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cm

Open ended pipes don't get painted on the ends, meaning the area of the two circles at both ends is not included in the calculation for surface area.

The surface area of the open-ended cylinder does not include the two circles: $\begin{align} SA & = A_{curved\ rectangle} + \cancel{A_{circles}} \\ \\ SA & = 2\pi r \times L + \cancel{2(\pi r^2)} \\ \\ SA & = 2\pi r \times L \end{align}$ Make sure to convert to the same units [5 cm = 0.05 m] Now substitute the known values into the equation, and solve, $\begin{align} SA & = 2\pi r \times L \\ \\ 23 & = 2\pi (0.05) \times L \\ \\ \dfrac{23}{2\pi (0.05)} & = \dfrac{\cancel{2\pi (0.05)}}{\cancel{2\pi (0.05)}} \times L \\ \\ L & = \dfrac{23}{2\pi (0.05)} \\ \\ & = 73.2\ cm \end{align}$ Therefore a gallon of paint can paint approximately 73.2 cm of that pipe.

Cylinder Surface Area

The circumference of an open-top cylindrical vase is 0.30 m. If the surface area (including the base) is 912 cm², determine the height 'h'. Solution

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cm

Hint Unavailable

First, determine the radius, given the circumference of the circle, where 0.30 m = 30 cm, $\begin{align} C & = 2\pi (r) \\ \\ 30 & = 2\pi (r) \\ \\ \dfrac{30}{2} & = \pi (r) \\ \\ 15 & = \pi r \\ \\ r & = \dfrac{15}{\pi}\ cm \\ \\ r & ≈ 4.78\ cm \end{align}$ Now see that the surface area of the open-top vase is the sum of one circle (the base) plus the rectangular curved area, $\begin{align} SA & = 1(Area_{circle}) + Area_{rectangular\ side} \\ \\ SA & = \pi (r)^2 + 2 \pi (r)(h) \\ \\ 912 & = \pi \left(4.78\right)^2 + 2 \pi \left(4.78\right)(h) \\ \\ 912 & = 71.74 + 30.02(h) \\ \\ 30.02(h) & = 912 - 71.74 \\ \\ 30.02(h) & = 840.26 \\ \\ h & = \dfrac{840.26}{30.02} \\ \\ h & = 28\ cm \end{align}$

Cylinder Volume

For the following cylinder, (not drawn to scale)

If the height is 15 mm and the radius is $\sqrt{3}$ cm, determine the volume, in cm³. Solution

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cm³

Hint Unavailable

First, convert 15 mm to 1.5 cm [There are 10 mm in 1 cm] $\begin{align} V & = (Area_{base})(height) \\ \\ & = (\pi r^2)(h) \\ \\ & = \pi (\sqrt{3})^2(1.5) \\ \\ & = 4.5\pi \\ \\ & = 14.1\ cm^3 \end{align}$

If the volume is 2π m³ and the height is 800 cm, determine the radius, in meters. Solution

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m

Hint Unavailable

First convert 800 cm to 8 m [There are 100 cm in 1 m... See that it is easier to convert cm to m, rather than convert m³ to cm³] $\begin{align} V & = (Area_{base})(height) \\ \\ V & = (\pi r^2)(h) \\ \\ 2\pi & = \pi (r)^2(8) \\ \\ 2\cancel{\pi} & = \cancel{\pi} (r)^2(8) \\ \\ 2 & = 8r^2 \\ \\ r^2 & = \dfrac{2}{8} \\ \\ r^2 & = \dfrac{1}{4} \\ \\ \sqrt{r^2} & = \sqrt{\dfrac{1}{4}} \\ \\ r & = 0.5\ m \end{align}$

Determine the radius 'r' if the volume is 375π cm³, and the height is triple the radius. Solution

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cm

Hint Unavailable

If the "the height is triple the radius", $\begin{align} height & = radius \times 3 \\ \\ h & = 3r \quad\quad\quad\quad [\text{Not} \ h = r^3] \end{align}$ Now use these values in the equation for volume of a cylinder, $\begin{align} V & = (Area_{base})(height) \\ \\ V & = (\pi r^2)(h) \\ \\ 375\pi & = (\pi r^2)({\color{DeepSkyBlue}3r}) \\ \\ 375\cancel{\pi} & = (\cancel{\pi} r^2)(3r) \\ \\ 375 & = (r^2)(3r) \\ \\ 375 & = (r)(r)(3)(r) \\ \\ 375 & = (3)(r)(r)(r) \\ \\ 375 & = (3)(r^3) \\ \\ r^3 & = \dfrac{375}{3} \\ \\ r^3 & = 125 \end{align}$ Now, depending on people's abilities, 'r' can be solved either with guess-and-check (left), or with cube roots (right)...
$\begin{align} r^3 & = 125 \\ \\ (\text{__})^3 & = 125 \\ \\ (5)^3 & = 125 \\ \\ r & = 5\ cm \end{align}$ $\begin{align} r^3 & = 125 \\ \\ \sqrt[3]{r^3} & = \sqrt[3]{125} \\ \\ r & = 5\ cm \end{align}$

Determine the height 'h' if the volume is 72π cm³, and the radius is triple the height. Solution

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cm

Hint Unavailable

If the "the radius is triple the height", $\begin{align} radius & = height \times 3 \\ \\ r & = 3h \quad\quad\quad\quad [\text{Not} \ r = h^3] \end{align}$ Now use these values in the equation for volume of a cylinder, $\begin{align} V & = (Area_{base})(height) \\ \\ V & = (\pi r^2)(h) \\ \\ 72\pi & = [\pi ({\color{DeepSkyBlue}3h})^2](h) \\ \\ 72\pi & = [\pi ({\color{DeepSkyBlue}3h})({\color{DeepSkyBlue}3h})](h) \\ \\ 72\pi & = [\pi (3)(3)(h)(h)](h) \\ \\ 72\pi & = \pi (9)(h^3) \\ \\ 72\cancel{\pi} & = \cancel{\pi} (9)(h^3) \\ \\ 72 & = 9 h^3 \\ \\ h^3 & = \dfrac{72}{9} \\ \\ h^3 & = 8 \end{align}$ Now, depending on people's abilities, 'r' can be solved either with guess-and-check (left), or with cube roots (right)...
$\begin{align} h^3 & = 8 \\ \\ (\text{__})^3 & = 8 \\ \\ (2)^3 & = 8 \\ \\ r & = 2\ cm \end{align}$ $\begin{align} r^3 & = 8 \\ \\ \sqrt[3]{r^3} & = \sqrt[3]{8} \\ \\ r & = 2\ cm \end{align}$

Optimization of Rectangle Area

A pet store owner has 90cm of fencing to enclose a baby turtle with three walls of fencing. Determine the dimensions of fencing that would allow for a maximum area of the enclosure for the baby turtle. Complete the table in your notebook, and show your work. Solution

Length (cm) Width (cm) Area (cm²)

1 _ _

2 _ _

3 _ _

5 _ _

6 _ _

9 _ _

10 _ _

15 _ _

18 _ _

30 _ _

45 _ _

The 30cm × 30cm = 900cm². $\begin{align} Width & = \dfrac{90 - 30}{2} \\ \\ & = \dfrac{60}{2} \\ \\ & = 30\ cm \\ \\ \\ \\ \\ \\ Area & = l × w \\ \\ & = 30\ cm × 30\ cm \\ \\ & = 900\ cm^2 \\ \\ \end{align}$ By inspection, you see that the 30cm × 30cm enclosure has the maximum area, 900cm².

Volume of Prisms

The formula shown is used to calculated the volume of: Solution V = (Area of base) × (Height)

Rectangular prism

Cylinder

Triangular prism

All of the above

The volume of a cube, rectangular prism, cylinder, and triangular prism are calculated with the formula: $\begin{align} Volume & = (Area\ of\ the\ base) × (Height) \\ \\ \\ \\ Volume_{rectangular} & = (L × W) × (Height) \\ \\ Volume_{cylinder} & = (\pi r^2) × (Height) \\ \\ Volume_{triangular} & = \left(\dfrac{1}{2}bh\right) × (Height) \\ \\ \end{align}$

Volume of Prisms

Determine the volume of the triangular prism below, where the height of the triangle is 5 cm, the base is 4 cm, and the height of the prism is 10cm. Solution

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cm³

Hint Unavailable

Volume = (Area of Base) × (Height) $\begin{align} Volume & = Area_{triangle} \ × \ Height_{prism} \\ \\ & = \dfrac{1}{2}b\cdot h \ × \ h_2 \\ \\ & = \dfrac{1}{2}(4)\cdot (5) \ × \ 10 \\ \\ & = \dfrac{1}{2}(20) \ × \ 10 \\ \\ & = 10 \ × \ 10 \\ \\ & = 100\ cm^3 \\ \\ \end{align}$

Surface Area of Prisms

The following triangular prism is open-topped and has an equilateral base. The perimeter of the base is 60 cm, and the surface area is 1,973.2 cm².

Calculate the height of the prism 'H₂'. Solution

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cm

Hint Unavailable

Calculate the length of each side of the triangular base, given its perimeter, $\begin{align} P & = 3(base) \\ \\ 60 & = 3b \\ \\ b & = \dfrac{60}{3} \\ \\ b & = 20\ cm \end{align}$ Find the height 'h' using pythagorean theorem. Use the right-angled triangle in the base with sides: 'h' 'b' and 'b/2' $\begin{align} b^2 & = h^2 + \left(\dfrac{b}{2}\right)^2 \\ \\ (20)^2 & = h^2 + \left(\dfrac{(20)}{2}\right)^2 \\ \\ 400 & = h^2 + 100 \\ \\ h^2 & = 400 - 100 \\ \\ h & = \sqrt{300}\ cm \\ \\ h & = 17.3\ cm \end{align}$ Given the surface area (one triangle plus three rectangles), you can find the height of the prism 'h₂' $\begin{align} SA & = 1 \times \left(Area_{triangle}\right) + 3 \times \left(Area_{rectangle}\right) \\ \\ SA & = 1 \times \left(\dfrac{1}{2}bh\right) + 3 \times \left(bh_2\right) \\ \\ 1,973.2 & = 1 \times \left(\dfrac{1}{2}(20)(17.3)\right) + 3 \times (20)(h_2) \\ \\ 1,973.2 & = 173 + 60\, h_2 \\ \\ 60\, h_2 & = 1,973.2 - 173 \\ \\ h_2 & = \dfrac{1,800.2}{60} \\ \\ h_2 & = 30\ cm \end{align}$

Determine the volume of the prism. Solution

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Calculate the length of each side of the triangular base, given its perimeter, $\begin{align} P & = 3(base) \\ \\ 60 & = 3b \\ \\ b & = \dfrac{60}{3} \\ \\ b & = 20\ cm \end{align}$ Find the height 'h' using pythagorean theorem. Use the right-angled triangle in the base with sides: 'h' 'b' and 'b/2' $\begin{align} b^2 & = h^2 + \left(\dfrac{b}{2}\right)^2 \\ \\ (20)^2 & = h^2 + \left(\dfrac{(20)}{2}\right)^2 \\ \\ 400 & = h^2 + 100 \\ \\ h^2 & = 400 - 100 \\ \\ h & = \sqrt{300}\ cm \\ \\ h & = 17.3\ cm \end{align}$ Given the surface area (one triangle plus three rectangles), you can find the height of the prism 'h₂' $\begin{align} SA & = 1 \times \left(Area_{triangle}\right) + 3 \times \left(Area_{rectangle}\right) \\ \\ SA & = 1 \times \left(\dfrac{1}{2}bh\right) + 3 \times \left(bh_2\right) \\ \\ 1,973.2 & = 1 \times \left(\dfrac{1}{2}(20)(17.3)\right) + 3 \times (20)(h_2) \\ \\ 1,973.2 & = 173 + 60\, h_2 \\ \\ 60\, h_2 & = 1,973.2 - 173 \\ \\ h_2 & = \dfrac{1,800.2}{60} \\ \\ h_2 & = 30\ cm \end{align}$ Now find volume of the prism, $\begin{align} Volume & = (Base\ Area)(Prism\ Height) \\ \\ & = \left(\dfrac{1}{2}bh\right)(H_2) \\ \\ & = \left(\dfrac{1}{2}(20)(17.3)\right)(30) \\ \\ & = 5,190\ cm^3 \end{align}$

Volume and Surface Area of Prisms

The height 'H' of the hexagonal prism is 3 cm, and the side length 's' is 6 cm. Determine the volume. Solution

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cm³

Hint Unavailable

First, determine the height of the equilateral triangle, using Pythagorean theorem, $\begin{align} H_{triangle} & = \sqrt{(side)^2 - \left(\dfrac{side}{2}\right)^2} \\ \\ & = \sqrt{(6)^2 - \left(\dfrac{6}{2}\right)^2} \\ \\ & = \sqrt{36 - 9} \\ \\ & = \sqrt{27}\ cm \end{align}$ Then, determine the area of the base, which consists of 6 equilateral triangles, $\begin{align} A_{base} & = 6 \times A_{triangle} \\ \\ & = 6 \times \left(\dfrac{1}{2}(base)(height)\right) \\ \\ & = 6 \times \left(\dfrac{1}{2}(6)(\sqrt{27})\right) \\ \\ & = 18\sqrt{27}\ cm^2 \end{align}$ Then, determine the volume of the prism, $\begin{align} V_{prism} & = A_{base} \times (Prism\ Height) \\ \\ & = (18\sqrt{27})(3) \\ \\ & = 54\sqrt{27}\ cm^3 \\ \\ & = 280.6\ cm^3 \end{align}$

The trough below is an open-top, equilateral, triangular prism. Given the volume of the triangular prism is 25,155.76 cm³, the surface area of the outside is 4,006.23 cm², and the length 'L' is 50 cm, determine the side length 's' of the triangle. Solution

s =

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cm

Based on what is given, you don't need height 'h' to solve for the side length 's'

For the triangular prism, the equation for volume is typical, but the surface area (SA) of the open-topped shape includes the ends plus two rectangular sides, instead of the usual three.
$\begin{align} Volume & = {\color{Red}Area_{triangle}} \ × \ Length \end{align}$ $\begin{align} SA & = 2({\color{Red}Area_{triangle}}) + 2(Area_{rectangle}) \end{align}$
First, you can find the area of the triangle, because you have the volume and length $\begin{align} Volume & = Area_{triangle} \ × \ Length \\ \\ Area_{triangle} & = \dfrac{Volume}{Length} \\ \\ Area_{triangle} & = \dfrac{25,155.76}{50} \\ \\ {\color{Red}Area_{triangle}} & = 503.1152\ cm^2 \end{align}$ Now that you have the area of the triangle, you can substitute this into the equation for the surface area, $\begin{align} SA & = 2({\color{Red}Area_{triangle}}) + 2(Area_{rectangle}) \\ \\ SA & = 2(503.1152\ cm^2) + 2(Length \ × \ Side) \\ \\ 4,006.23 & = 2(503.1152\ cm^2) + 2(50 \ × \ Side) \\ \\ 4,006.23 & = 1,006.2304 + 100 \ × \ Side \\ \\ 4,006.23 - 1,006.2304 & = 100 \ × \ Side \\ \\ 2,999.9996 & = 100 \ × \ Side \\ \\ Side & = \dfrac{2,999.9996}{100} \\ \\ Side & = 30\ cm \end{align}$

Surface Area with Compound Shapes

Given the equations,

$\begin{align} \text{Area}_{\, parallelogram} = (base)(height) \\ \\ \text{Area}_{\, trapezoid} = \dfrac{1}{2}(height)(top + base) \end{align}$

Determine the area of the shaded part of the figure below, if the area of the parallelogram is 66 units². Solution

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Method 1: Determine the base of the parallelogram, $\begin{align} \text{Area}_{\, parallelogram} & = (base)(height) \\ \\ 66 & = (b)(11) \\ \\ b & = \dfrac{66}{11} \\ \\ b & = 6 \end{align}$ Then calculate the area of the whole (larger) trapezoid, $\begin{align} \text{Area}_{\, trapezoid} & = \dfrac{1}{2}(11)[12 + (14 + 6)] \\ \\ & = 176 \end{align}$ Then subtract the area of the parallelogram from the area of the trapezoid, $\begin{align} \text{Area}_{\, shaded} & = \text{Area}_{\, trapezoid} - \text{Area}_{\, parallelogram} \\ \\ & = 176 - 66 \\ \\ & = 110\ units^2 \end{align}$ Method 2: Determine the base of the parallelogram, $\begin{align} \text{Area}_{\, parallelogram} & = (base)(height) \\ \\ 66 & = (b)(11) \\ \\ b & = \dfrac{66}{11} \\ \\ b & = 6 \end{align}$ Then, determine the top of the shaded trapezoid, $\begin{align} top & = 12 - 6 \\ \\ & = 6 \end{align}$ Then calculate the area of the shaded trapezoid trapezoid, $\begin{align} \text{Area}_{\, trapezoid} & = \dfrac{1}{2}(11)[6 + 14] \\ \\ & = 110\ units^2 \end{align}$

Determine the ratio of the shaded trapezoid area to the whole trapezoid area. Reduce fully. Solution

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:

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Earlier, you determined the area of...

Then calculate the area of the whole (larger) trapezoid, $\begin{align} \text{Area}_{\, trapezoid} & = \dfrac{1}{2}(11)[12 + (14 + 6)] \\ \\ & = 176 \end{align}$ Then subtract the area of the parallelogram from the area of the trapezoid, $\begin{align} \text{Area}_{\, shaded} & = \text{Area}_{\, trapezoid} - \text{Area}_{\, parallelogram} \\ \\ & = 176 - 66 \\ \\ & = 110\ units^2 \end{align}$ So the ratio of the shaded trapezoid area to the whole trapezoid area is... [And remember to reduce fully!] $\begin{align} Area_{shaded} & : Area_{whole} \\ \\ 110 & : 176 \\ \\ 55 & : 88 \\ \\ 5 & : 8 \end{align}$

Volume of Prisms with Compound Shapes

Remember that for prisms with constant cross-sectional areas, the volume can be calculated with the product of the area of the base (side), and the length (or height).

$\begin{align} & \text{Volume}_{\, prism} = (area\ of\ base)(length) \\ \\ & \text{Volume}_{\, parallelogram\ prism} = (base)(height)(length) \\ \\ & \text{Volume}_{\, triangular\ prism} = \dfrac{1}{2}(base)(height)(length) \\ \\ & \text{Volume}_{\, trapezoidal\ prism} = \dfrac{1}{2}(height)(top + base)(length) \end{align}$

Given the volume of the trapezoidal prism is 1,000 cm², the top 'a' is 5 cm, base 'b' is 11 cm, and length 'L' is 10 cm, determine the height 'h'. Solution

h =

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cm

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$\begin{align} \text{Volume}_{\, prism} & = (area\ of\ base)(length) \\ \\ \text{Volume}_{\, trapezoidal\ prism} & = \dfrac{1}{2}(height)(top + base)(length) \\ \\ 1,000 & = \dfrac{1}{2}(height)(5 + 11)(10) \\ \\ 1,000 & = \dfrac{1}{2}(height)(16)(10) \\ \\ 1,000 & = 180(height) \\ \\ \dfrac{1,000}{180} & = (height) \\ \\ h & = 12.5\ cm \\ \\ \end{align}$

Calculate the volume of the following hexagonal prism, given: height 'H' is 12 cm, side 's' 7 cm, area of the shaded triangle is 36 cm², and height of the triangle 'h' is 5 cm. Solution

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Given the area of the shaded triangle, determine the base 'b' $\begin{align} \text{Area}_{\, triangle} & = \dfrac{1}{2}(base)(height) \\ \\ 36 & = \dfrac{1}{2}(base)(5) \\ \\ 36 & = 2.5(base) \\ \\ b & = \dfrac{36}{2.5} \\ \\ b & = 14.4\ cm \end{align}$ Now you can calculate the area of the whole base of the hexagonal prism. See that a hexagon consists of compound shapes: two triangles, plus a rectangle, $\begin{align} \text{Area}_{\, hexagon} & = 2(\text{Area}_{\, triangle}) + \text{Area}_{\, rectangle} \\ \\ & = 2(36) + (14.4)(7) \\ \\ & = 172.8\ cm^2 \end{align}$ Now that you have everything you need, calculate the volume of the hexagonal prism, $\begin{align} \text{Volume}_{\, prism} & = (area\ of\ base)(Height) \\ \\ & = (\text{Area}_{\, hexagon})(Height) \\ \\ & = (172.8\ cm^2)(12\ cm) \\ \\ & = 2,073.6\ cm^3 \end{align}$

Calculate the height 'h' in the figure below, given the base 'b' is 8 cm, the length 'L' is 10 cm, the total volume is 1,250 cm³, and the height 'h' of the triangular prism, is equal to that of the parallelogram prism. Solution

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$\begin{align} \text{Volume}_{\, Total} & = \text{Volume}_{\, triangular\ prism} + \text{Volume}_{\, parallelogram\ prism} \\ \\ 1250 & = \dfrac{1}{2}(b)(h)(L) + (b)(h)(L) \\ \\ 1250 & = \dfrac{1}{2}(8)(h)(10) + (8)(h)(10) \\ \\ 1250 & = 40(h) + 80(h) \\ \\ 1250 & = 120(h) \\ \\ h & = \dfrac{1250}{120} \\ \\ h & = 10.42 cm \end{align}$

Angles, Coordinates, Geometry

Shapes

Polygons

What is a "regular polygon"? Provide at least one example. [3] Solution

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Any two of the following with same sides and angles: equilateral triangle, square, pentagon, hexagon, ...

Regular polygons have all side lengths and angles equal.

Polygons

Which of the following polygons is not 4-sided? Solution

Quadrilateral

Rectangle

Rhombus

Square

Trapezoid

Parallelogram

Kite

All of the above are 4-sided

All are 4-sided:

Quadrilateral

Rectangle

Rhombus

Square

Trapezoid

Parallelogram

Kite

Quadrilateral Polygon Classification Exercise

Given the 8 terms, arrange into the categories or hierarchy - "what falls under what"...? Solution

Square (all sides equal, all right angles)

Quadrilateral (4 sides)

Rectangle (all right angles)

Polygon (closed, 2D, straight sides)

Rhombus (all sides equal)

Parallelogram (2 parallel sides)

Kite (two different pairs of equal, adjacent sides)

Trapezoid (at least one pair of parallel sides)

This is the general order... Google 'em!

Classifying Shapes

A rhombus is a parallelogram with equal side lengths, including squares and rectangles. Solution

True

False

A rhombus is a parallelogram with equal side lengths.

However this can only include squares.

Applications of Geometric Properties

Which of the following shapes has the greatest area/perimeter ratio? Solution

Square

Parallelogram

Equilateral Triangle

Any quadrilateral

Circle

The circle has the greatest ratio of: $\dfrac{surface\ area}{perimeter}$ [Calculations not shown]

Angles

Interior Angle Notation

Which of the following represents $\angle$ BAC ? Solution

I

II

III

None of the above

$\angle$ BAC references an angle in the triangle with the middle letter, 'A' representing the angle, I.

Transversal Angles

Which of the following angles represent corresponding angles? Solution

$\angle$ A and $\angle$ D

$\angle$ D and $\angle$ Z

$\angle$ C and $\angle$ W

$\angle$ C and $\angle$ X

Corresponding angles are also known as the 'F' angles.

Other corresponding angles are $\angle$ C and $\angle$ Y

or $\angle$ A and $\angle$ W...

Which of the following angles represent alternate angles? Solution

$\angle$ W and $\angle$ Z

$\angle$ B and $\angle$ Y

$\angle$ D and $\angle$ W

$\angle$ A and $\angle$ Z

Alternate angles are also known as the 'Z' angles.

Other alternate angles are $\angle$ C and $\angle$ X.

Angles on a Line

Determine the angles on each side below. Solution

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The sum of angles on a line = 180˚ $\begin{align} \sum \text{Angles} & = 180˚ \\ \\ \text{Angles}_1 + \text{Angles}_2 & = 180˚ \\ \\ (x - 4) + (2x + 1) & = 180˚ \\ \\ 3x - 3 & = 180˚ \\ \\ 3x & = 180˚ + 3 \\ \\ 3x & = 183˚ \\ \\ x & = \dfrac{183˚}{3} \\ \\ x & = 61˚ \\ \\ x - 4 & = 61˚ - 4 = \underline{57˚} \\ \\ 2x + 1 & = 2(61˚) + 1 = \underline{123˚} \\ \\ \end{align}$

Pythagorean Theorem

The pythagorean theorem states the: Solution

Sum of the squares of the sides of a triangle is 50.

The hypotenuse is 1 more than the longest side length.

The square of the hypotenuse of all triangles is equal to the sum of the squares of the shorter sides.

The square of the hypotenuse of a right triangle is equal to the sum of the squares of the shorter sides.

As you can see in the diagram, the area of the square on the longest side (hypotenuse) is 5² = 25.
The sum of the areas of the squares on the two shorter sides is 3² + 4² = 9 + 16 = 25.

∴ The area of the larger square is always equal to the sum of the two smaller areas.

Pythagorean Theorem

Which of the following is the correct formula to solve for side length b in a right triangle using the Pythagorean theorem? Solution

$b = \sqrt{c^2 + a^2}$

$b = \sqrt{a^2 - c^2}$

$b = \sqrt{c^2 - a^2}$

$b = \sqrt{a^2 × c^2}$

$\begin{align} a^2 + b^2 & = c^2 \\ \\ b^2 & = c^2 - a^2 \\ \\ \sqrt{b^2} & = \sqrt{c^2 - a^2} \\ \\ \sqrt{b^\cancel{2}} & = \sqrt{c^2 - a^2} \\ \\ b & = \sqrt{c^2 - a^2} \\ \\ \end{align}$

Pythagorean Theorem

Two right triangles share a side length 'c' in the diagram below.

Determine the length of the hypotenuse 'c'. Solution

c =

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$\begin{align} a^2 + b^2 & = c^2 \\ \\ 10^2 + 8^2 & = c^2 \\ \\ 100 + 64 & = c^2 \\ \\ \sqrt{100 + 64} & = \sqrt{c^\cancel{2}} \\ \\ \sqrt{100 + 64} & = c \\ \\ c & = 12.8 \\ \\ \end{align}$

Determine the length of side 'z'. Solution

z =

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c = 12.8 $\begin{align} c^2 + 3^2 & = z^2 \\ \\ 12.8^2 + 3^2 & = z^2 \\ \\ 164 + 9 & = z^2 \\ \\ \sqrt{164 + 9} & = \sqrt{z^\cancel{2}} \\ \\ \sqrt{164 + 9} & = z \\ \\ z & = 13.2 \\ \\ \end{align}$

Coordinates

Coordinate Translations

Determine the coordinate of the image formed from the given point (x, y) with the shift/slide by the translation vector [h, k]. Otherwise determine whatever else is asked.

Left: (x - h, y)
Right: (x + h, y)
Up: (x, y + k)
Down: (x, y - k)

Point: (2, 3) with the translation vector: [1, 1] Solution

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( , )

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$\begin{align} & = (2 + 1, 3 + 1) \\ \\ & = (3, 4) \end{align}$

Point: (-4, 5) with the translation vector: [3, -7] Solution

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( , )

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$\begin{align} & = (-4 + 3, 5 - 7) \\ \\ & = (-1, -2) \end{align}$

Point: (-11, -15) with the translation vector: [14, 20] Solution

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( , )

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$\begin{align} & = (-11 + 14, -15 + 20) \\ \\ & = (3, 5) \end{align}$

Determine the original coordinate, given the image (5, -2) and the translation vector: [-12, 2] Solution

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( , )

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$\begin{align} (5, -2) & = [\text{_} - 12, \text{_} + 2] \\ \\ (5, -2) & = (17 - 12, -4 + 2) \end{align}$ So, the original point is (17, -4)

Coordinate Dilatations/Resizing

Determine the coordinate of the image formed from the given point (x, y) with the dilatation by the scale factor 'a'. Otherwise determine whatever else is asked.

a(x, y) = (a × x, a × y)

Scale (0, 0) by a factor of 2 Solution

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( , )

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$\begin{align} & = (0, 0) \\ \\ & = (0 × 2, 0 × 2) \\ \\ & = (0, 0) \end{align}$

Scale (2, -6) by a factor of 5 Solution

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( , )

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$\begin{align} & = (2, -6) \\ \\ & = (2 × 5, -6 × 5) \\ \\ & = (10, -30) \end{align}$

Scale (6, 12) by a factor of $\dfrac{1}{2}$ Solution

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( , )

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$\begin{align} & = (6, 12) \\ \\ & = \left(6 \times \dfrac{1}{2}, 12 \times \dfrac{1}{2}\right) \\ \\ & = \left(\dfrac{6}{2}, \dfrac{12}{2}\right) \\ \\ & = (3, 6) \end{align}$

Scale (-3, 0) by a factor of $\dfrac{1}{3}$ Solution

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( , )

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$\begin{align} & = (-3, 0) \\ \\ & = \left(-3 \times \dfrac{1}{3}, 0 \times \dfrac{1}{3}\right) \\ \\ & = \left(\dfrac{-3}{3}, \dfrac{0}{3}\right) \\ \\ & = (-1, 0) \end{align}$

Coordinate Reflections/Mirroring

Determine the coordinate of the image formed from the given point (x, y) with the reflection across the different axes. Otherwise determine whatever else is asked.

Mirror across the 'x'-axis: (x, y) → (x, -y)
Mirror across the 'y'-axis: (x, y) → (-x, y)
Mirror across the line 'y = x': (x, y) → (y, x)

Reflect (3, 2) across the x-axis Solution

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( , )

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$\begin{align} & = (3, 2) \\ \\ & = (3, -2) \end{align}$

Reflect (-2, 0) across the y-axis Solution

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( , )

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$\begin{align} & = (-2, 0) \\ \\ & = (-(-2), 0) \\ \\ & = (2, 0) \end{align}$

Reflect (3, -4) across the line, 'y = x' Solution

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( , )

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$\begin{align} & = (3, -4) \\ \\ & = (-4, -3) \end{align}$

Reflect (-2, -1) across the line, 'y = x' Solution

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( , )

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$\begin{align} & = (-2, -1) \\ \\ & = (-1, -2) \end{align}$

Coordinate Rotations

Determine the coordinate of the image formed from the given point (x, y) with the rotation about the origin (0, 0). Otherwise determine whatever else is asked.

(x, y)

Rotation	90˚	180˚	270˚
Clockwise (CW)	(y, -x)	(-x, -y)	(-y, x)
Counterclockwise (CCW)	(-y, x)	(-x, -y)	(y, -x)

A 90˚ CW rotation is the same as a 270˚ CCW rotation. Solution

True

False

True, see the table above.

A 90˚ CCW rotation is the same as a 270˚ CW rotation. Solution

True

False

True, see the table above.

A 180˚ CW rotation is the same as a 180˚ CCW rotation. Solution

True

False

True, see the table above.

Rotate (1, 3) 90˚ CCW about the origin. Solution

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( , )

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$\begin{align} & (1, 3) \\ \\ & = (-3, 1) \end{align}$

Rotate (-2, 3) 180˚ CW about the origin. Solution

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( , )

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$\begin{align} & (-2, 3) \\ \\ & = (-(-2), -3) \\ \\ & = (2, -3) \end{align}$

Rotate (0, -5) 270˚ CW about the origin. Solution

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( , )

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$\begin{align} & = (0, -5) \\ \\ & = (-5, 0) \end{align}$

Rotate (a, b) 90˚ CW about the origin. Solution

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( , )

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$\begin{align} & (a, b) \\ \\ & = (b, a) \end{align}$

Coordinate Movement Combinations

Determine the transformations that will move the solid parallelogram overtop the dotted location. Original coordinates: A(0, 4), B(3, 7), C(10, 7), and D(7, 4). Solution

Rotate 90˚ clockwise about the origin, 7 right, 3 up

Reflect across the x-axis, rotate 180˚ clockwise about the origin, 11 down, 14 right

Rotate 90˚ counterclockwise about the origin, 18 right, 7 down

Any of the above

Any of the above. [coordinate translation/rotation work not shown]

Relationship Between Two Variables

Complete the following table relating the side length of a square to its area. Solution

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Side Length (cm) Area (cm²)

2 4

5 25

7

10

144

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Side Length (cm) Area (cm²)

2 4

5 25

7 49

10 100

12 144

Graphing

Graph the data in the following table on a scatter plot. Make sure to properly label and set the dependent and independent variables on the correct axis. Solution

Number of White Roses Cost of Flower Arrangement

1 $3.00

4 $12.00

8 $24.00

12 $36.00

16 $48.00

24 $65.00

Make points on the axis that have equally spaced intervals, like 5, 10, 15, 20, 25, 30, etc... Don't use 1, 4, 8, 12, 16, 24 on your x-axis because those points have uneven spacing!

Make sure to label the axes and make a title of the graph.

Describe the trend in your relation. Solution

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Know the terms,

Strong vs. Weak

Positive vs. Negative

The trend is strong because the points appear to hover near a line of best fit. The trend is positive because as the independent axis increases in value, the dependent axis increases in value.

Estimate the cost for 10 roses. Solution

$

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~$30

Relations in Tables

Given the equation:

$\begin{align} y & = -5x + 3 \end{align}$

Complete the table of coordinates using the line equation determined previously. Solution

Hint Clear Info

x y

-2

-1

0

1

2

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Substitute the 'x'-values into the equation, to determine the outputs ('y'-values)...

x y

-2 -5(-2) + 3 = 10 + 3 = 13

-1 -5(-1) + 3 = 5 + 3 = 8

0 -5(0) + 3 = 0 + 3 = 3

1 -5(1) + 3 = -5 + 3 = -2

2 -5(2) + 3 = -10 + 3 = -7

Linear Equation Algebra

A car-rental company charges a monthly fee of $25 plus $8 per hour, n of use for a car.

Write an equation for the cost, C. Solution

C =

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Notice that this is an equation in the form y = mx + b ...

slope (m) is the hourly rate, and the y-intercept (b) is the fixed amount.
$\begin{align} C & = (\text{Hourly rate})(\text{Number of hours}) + \text{Fixed amount} \\ \\ C & = 8(n) + 25 \\ \\ \end{align}$

Rearrange this equation for 'n' and determine how many hours the car can be used for $465. Solution

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hours

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C = $465 $\begin{align} C & = 25 + 8(n) \\ \\ C - 25 & = 8n \\ \\ 8n & = C - 25 \\ \\ 8n & = 465 - 25 \\ \\ 8n & = 440 \\ \\ \frac{\cancel{8}n}{\cancel{8}} & = \frac{440}{8} \\ \\ n & = \frac{440}{8} \\ \\ n & = 55\ hours \\ \\ \end{align}$

If a different car rental company charges no monthly cost with a higher rate of $12 per hour, determine which company is a better deal for the same amount of use. Solution

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$\begin{align} C & = 12(n) + 0 \\ \\ C & = 12n \\ \\ C & = 12(55) \\ \\ C & = \$660 \\ \\ \end{align}$ The cost is higher, $660 for the same amount of hours, so it is not a better deal.

Relations in Equations

Determine the missing value in the coordinate, for the equation:

12x - 3y = 36

Solution

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(4, )

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$\begin{align} 12x - 3y & = 36 \\ \\ -3y & = - 12x + 36 \\ \\ \dfrac{-3y}{-3} & = - \dfrac{12x}{-3} + \dfrac{36}{-3} \\ \\ y & = 4x - 12 \\ \\ y & = 4(4) - 12 \\ \\ y & = 16 - 12 \\ \\ y & = 4 \end{align}$

Solution

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( , 20)

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$\begin{align} 12x - 3y & = 36 \\ \\ -3y & = - 12x + 36 \\ \\ \dfrac{-3y}{-3} & = - \dfrac{12x}{-3} + \dfrac{36}{-3} \\ \\ y & = 4x - 12 \\ \\ 20 & = 4x - 12 \\ \\ 20 + 12 & = 4x \\ \\ 4x & = 32 \\ \\ \dfrac{4x}{4} & = \dfrac{32}{4} \\ \\ x & = 8 \end{align}$

Coordinate Measurement

Given the following coordinates: A(2, 4), B(8, 4), C(6, -2), D(0, -2)...

Determine the perimeter of the four-sided figure. Round your answer to the nearest decimal. Solution

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units

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Length AB: $\begin{align} & = 8 - 2 \\ \\ & = 6\ units \end{align}$ Length DC: $\begin{align} & = 6 - 0 \\ \\ & = 6\ units \end{align}$ Use Pythagorean theorem to determine the slant lengths... Using Pythagorean theorem, length AD: $\begin{align} AD^2 & = (2-0)^2 + (-2-4)^2 \\ \\ AD^2 & = (2)^2 + (-6)^2 \\ \\ AD & = \sqrt{(2)^2 + (-6)^2} \\ \\ AD & = \sqrt{4 + 36} \\ \\ AD & = \sqrt{40} \\ \\ AD & = 6.3\ units \end{align}$ Using Pythagorean theorem, length BC: $\begin{align} BC^2 & = (8-6)^2 + (-2-4)^2 \\ \\ BC^2 & = (2)^2 + (-6)^2 \\ \\ BC & = \sqrt{(2)^2 + (-6)^2} \\ \\ BC & = \sqrt{4 + 36} \\ \\ BC & = \sqrt{40} \\ \\ BC & = 6.3\ units \end{align}$ So, the perimeter is: $\begin{align} & = AB + DC + AD + BC \\ \\ & = 6 + 6 + 6.3 + 6.3 \\ \\ & = 24.6\ units \end{align}$

Dilate the figure by a scale factor of 2, and calculate the new area. Solution

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units²

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Scale the 'x' and 'y' values by a factor of 2... Base length: $\begin{align} & = 12 - 0 \\ \\ & = 12\ units \end{align}$ Height: $\begin{align} & = 8 - (-4) \\ \\ & = 8 + 4 \\ \\ & = 12 \end{align}$ Calculate, $\begin{align} Area & = (base)(height) \\ \\ & = (12)(12) \\ \\ & = 144\ units^2 \end{align}$

The perimeter and area of the new figure area the same, for any rotation or reflection. Solution

True

False

True, the perimeter and area does not change for rotations or reflections.

Coordinate Applications

A triangle with 21 unit² area has points at A(4, -1) and B(10, -1). Determine a possible 'y'-value of the coordinate C. Solution

y =

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The 'base' length of the triangle can be calculated with, $\begin{align} base & = x_2 - x_1 \\ \\ & = 10 - 4 \\ \\ & = 6\ units \end{align}$ Then substitute the givens into the equation for area of a triangle, $\begin{align} Area & = \dfrac{1}{2}(base)(height) \\ \\ 21 & = \dfrac{1}{2}(6)(height) \\ \\ 21 & = 3(height) \\ \\ height & = \dfrac{21}{3} \\ \\ height & = 7\ units \end{align}$ Then the 'y'-value could be, either above or below the 'base' at y = -1, $\begin{align} y & = -1 \pm 7 \\ \\ & = -1 + 7,\ \text{or}\ -1 - 7 \\ \\ & = 6,\ \text{or}\ -8 \end{align}$ The point 'C' could be anywhere on the lines y = -8, or y = 6. For instance C(1, 6)

Coordinate Applications with Dilatations/Resizing

Determine the new dimension(s) for each of the following.

A figure with an area of 80 units² is drawn by connecting the coordinates A(4, 6), B(14, 6), C(12, -2), D(4, -2). Determine the new area after the coordinates are scaled by a factor of a half. Solution

Area =

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u²

Hint Unavailable

You can calculate the new area given the original area and the scale factors [not shown]. Or, you can calculate the new area using the scaled coordinates [shown]. Given that the original area (base × height) is 80 units², and the base and height are each scaled by a factor of a half... $\begin{align} Area_{\,New} & = \left(\dfrac{1}{2} \times base\right) \left(\dfrac{1}{2} \times height\right) \\ \\ & = \left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right) \left(base \times height\right) \\ \\ & = \left(\dfrac{1}{4}\right) \left(base \times height\right) \\ \\ & = \left(\dfrac{1}{4}\right) \left(80\ units^2\right) \\ \\ & = 20\ units^2 \end{align}$

A figure with vertices at A(0, 0), B(4, 0), C(4, 2), D(0, 2) is dilated by a factor of 3. The original area is 8 units², and the original perimeter is 12 units. Solution

Hint Clear Info

Area = u²
Perimeter = u

Incorrect Attempts:

CHECK

Hint Unavailable

I) You can calculate the new area given the original area and the scale factors $\begin{align} Area_{\,New} & = (3 \times base)(3 \times height) \\ \\ & = (3)(3)(base)(height) \\ \\ & = 9(base)(height) \\ \\ & = 9(8\ units^2) \\ \\ & = 72\ units^2 \end{align}$ II) Or, you can calculate the new area using the scaled coordinates

(0, 0) --> (0 × 3, 0 × 3) --> (0, 0)

(4, 0) --> (4 × 3, 0 × 3) --> (12, 0)

(4, 2) --> (4 × 3, 2 × 3) --> (12, 6)

(0, 2) --> (0 × 3, 2 × 3) --> (0, 6)

$\begin{align} Area_{\,New} & = (base)(height) \\ \\ & = (12 - 0)(6 - 0) \\ \\ & = (12)(6) \\ \\ & = 72\ units^2 \end{align}$ III) You can calculate the new perimeter using the original perimeter and scale factors [not shown]. Or, you can calculate the new perimeter using the scaled coordinates [shown below] $\begin{align} Perimeter_{\,New} & = 2\Big[base + height\Big] \\ \\ & = 2\Big[(12 - 0) + (6 - 0)\Big] \\ \\ & = 2\Big[18\Big] \\ \\ & = 36\ units \end{align}$

Patterning and Algebra

Patterning

Determine the missing terms, indicated by the symbol, █.

12, 24, 36, 48, █, 72, ...

50

54

60

62

+12

10, 14, 22, 34, 50, █, ...

70

68

64

54

+ 4n

35, 32, 26, 17, 5, █, ...

1

0

-5

-10

- 3n

Representing Pattern Rules from Words

Make a table of values and a graph for the following pattern rule. Plot the coordinates as the term number and the term. Solution Start with 2, then add one to each term and double it to get the next term.

x y

1 2

2 6

3 14

4 30

5 62

[Graph not shown]

Patterning with Multiplication and Division

Determine the pattern rule for the following pattern, without the use of a calculator. Solution Video 3, 12, 48, 192, 768...

Start with three then multiply by 3 to get the next term

Start with three then multiply the term number by 3 to get the next term

Start with three then multiply by 4 to get the next term

Start with three then multiply the term number by 4 to get the next term

Find out what it multiplies by each time: × 4. The pattern rule is multiply by 4...

Determine the 60th term, given the pattern t = 3n - 4 Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

n = 60 $\begin{align} t & = 3n - 4 \\ \\ & = 3(60) - 4 \\ \\ & = 176 \end{align}$

Formulas with Single Degree Variables

Solve the following.

If the area of a soccer field is 8,000m² and the width is 100m, find the length of the field. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

$\begin{align} area & = (length)(width) \\ \\ length & = \dfrac{area}{width} \\ \\ & = \dfrac{8,000m^2}{100m} \\ \\ & = 80\ m \\ \\ \end{align}$

The circumference of a circle is 29.83cm. What is the radius? Solution $C = 2\pi r$

Hint Clear Info

Incorrect Attempts:

CHECK

cm

Hint Unavailable

$\begin{align} C & = 2\pi r \\ \\ r & = \dfrac{C}{2\pi} \\ \\ & = \dfrac{29.83cm}{2(3.14)} \\ \\ & = 4.75\ cm \\ \\ \end{align}$

Determine the length of a rectangle with a perimeter of 37m and a width of 8.5m. Solution p = 2(length) + 2(width)

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

$\begin{align} perimeter & = 2(length) + 2(width) \\ \\ 37m & = 2(length) + 2(8.5m) \\ \\ 37m & = 2(length) + 17m \\ \\ 2(length) & = 37m - 17m \\ \\ 2(length) & = 20m \\ \\ length & = \dfrac{20m}{2} \\ \\ & = 10\ m \\ \\ \end{align}$

If the area of a ginormous triangle is 50m² and the height is 5m, determine the dimension of its base. Solution $a = \dfrac{bh}{2}$

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

$\begin{align} a & = \dfrac{bh}{2} \\ \\ 2a & = bh \\ \\ b & = \dfrac{2a}{h} \\ \\ & = \dfrac{2(50m^2)}{5m} \\ \\ & = 20\ m \\ \\ \end{align}$

If a store sells its slow-cooked meat sandwiches for $9.50 each and made $2,156.50 on sandwiches during the whole day, how many sandwiches were sold? Solution

Hint Clear Info

Incorrect Attempts:

CHECK

sandwiches

Hint Unavailable

$\begin{align} & = \dfrac{$2,156.50}{$9.50} \\ \\ & = 227\ sandwiches \\ \\ \end{align}$

Representing Situations with Algebra

Determine the correct equation or expression that represents the cost of pizza for a whole school, at $3.50 per student plus a $45 delivery fee. Solution

C = 3.5n

C = 3.5 + n

C = 45n

C = 45 + n

C = 3.5n + 45

Cost = $3.50(number) + $45

Evaluating Algebraic Expressions Up to 3 Terms

Evaluate.

24 + 3y - 2x, where $y = \dfrac{4}{3}$ , and x = 3. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = 24 + 3y - 2x \\ \\ & = 24 + 3\left(\dfrac{4}{3}\right) - 2(3) \\ \\ & = 24 + 4 - 6 \end{align}$

3(3x + y) - z, where x = 3.5, y = 0.5, z = 2 Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = 3(3x + y) - z \\ \\ & = 3(3(3.5) + 0.5) - 2 \end{align}$

Solving Algebra Equations With Integers and Single Variables

Solve for the variable in the following equations, show your work. Use any of the following methods: inspection, guess-and-check, or "balance"

25x - 5 = 45 Solution

x =

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

12y ÷ 3 = 8 Solution

y =

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

2x - 10 = x + 6

x =

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Data Management and Probability

Selecting the Appropriate Type of Graph

For each of the data sets below, determine the type of graph that best represents the data.

To indicate the future numbers of global population growth. Solution

Scatter Plot

Histogram

Bar graph

Circle (pie) graph

Conventionally, these increases in numbers are shown with a scatter plot, with time on the x-axis, and total numbers on the y-axis. The trend can be extended to predict population in future years using a line of best fit.

To compare the fraction of each type of web browser used by visitors of a website. Solution

Scatter Plot

Histogram

Bar graph

Circle (pie) graph

These relative, fraction or percentage comparisons are best displayed with a circle (a.k.a. pie) graph.

To show the brands of cars in a parking lot. Solution

Scatter Plot

Histogram

Bar graph

Circle (pie) graph

Typically, a bar chart would be used to show the number (on the y-axis) of each category (on the x-axis). A histogram would not commonly be used because histograms must have sequential numerical values on the x-axis.

To record the heights of all the students in your math class. Solution

Scatter Plot

Histogram

Bar graph

Circle (pie) graph

Conventionally, a histogram is used for this, with sequential intervals on the x-axis, and frequency count on the y-axis. Histograms have number values on the x-axis (vs. bar graphs that can have numbers, or category names).

Census Terminology

A sample is a larger representation than a population. Solution

True

False

Population includes the entire set of a study, while a sample is the subset of the population used to represent the population. A larger sample size is more representative of the entire population.

Explain why a sample is used to represent a population. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Sample are used to extrapolate and understand trends about populations. It is far less expensive, and less time consuming to collect data from a smaller sample, rather than the entire population. Often, if the sample is chosen well, it is a good representation of the properties of a population.

Measures of Central Tendency - Pizza Day

A highly diplomatic school is scheduled to have a pizza day, so it surveys its population for their preferred toppings. The school can only order one kind of pizza due to the large volume of the order. The vegetarian students get salads, not included here.

What measure of central tendency would be used to determine the pizza to order? Solution

Mean

Median

Mode

Any of the above

Mode is the measurement of the 'most common' data point, in this case the type of pizza that most students want, should be ordered.

What pizza should be ordered? Solution

Homeroom Classroom Pepperoni BBQ Chicken No Preference

7A 10 12 8

8D 11 15 5

9F 9 9 9

9C 6 8 10

10G 13 13 0

10B 14 15 1

11T 21 1 3

11P 6 16 6

12G 4 8 14

12D 3 11 18

BBQ Chicken

Pepperoni

Out of the 269 responses, 97 want pepperoni, and 108 want BBQ chicken.

Calculating Measures of Central Tendency

Given the data set representing the number of children, in the families that have children, on one particular street:

2, 2, 4, 1, 2, 3, 4, 6, 2, 3, 2, 3, 2

Determine the mean. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Mean is the average

$\begin{align} & = \dfrac{Sum}{total\ number} \\ \\ & = \dfrac{36}{13} \\ \\ & = 2.8 \end{align}$

Determine the median. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Median is the middle of the ordered data set

Order ascending: 1 2 2 2 2 2 2 3 3 3 For an odd number of values, the median is the middle value. For an even number of values, the median is the mean (average) of the middle two, which just equals 2.

Determine the mode. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

You don't need to order the numbers, but it makes it easier to pick out the most common value, the mode. 1 2 2 2 2 2 2 3 3 3 The most common value, the mode, is 2.

Calculating Measures of Central Tendency in Reverse

Given the rankings (scores) of players on a team in a video game, determine the rank of the missing player. The average ranking of the team is 1612. Solution 2827, 2213, 1670, 1478, ____, 1185, 1108, 1052

Hint Clear Info

Incorrect Attempts:

CHECK

You know the equation for calculating the average. In this case you can work backwards, because you are given two values (average, and number of values), and you can calculate the third (the missing ranking).

Depending on your ability (some people are still just learning algebra), you can solve this with guess-and-check or algebraic [shown here] methods... Let 'x' be the missing ranking. $\begin{align} average & = \dfrac{sum}{number\ of\ values} \\ \\ 1612 & = \dfrac{12,896 + x}{8} \\ \\ 1612(8) & = 12,896 + x \\ \\ 12,896 + x & = 1612(8) \\ \\ x & = 1612(8) - 12,896 \\ \\ x & = 1363 \\ \\ \end{align}$

Calculating Measures of Central Tendency - School Marks

A grade 12 student has the following final marks: 88%, 72%, 83%, 85%, and 86%

The student is applying to university and would like to receive the $2,000 entrance scholarship for an overall average above 90%. Is it possible for this student to receive the scholarship? Solution

Yes

No

The quickest way to check is to assume the highest score (100%) for the sixth class, and see what their average would be. Of course you can also work backwards, calculating the data point, given 6 terms, and the 90% average [not shown]. $\begin{align} average & = \dfrac{88 + 72 + 83 + 85 + 86 + {\color{DeepSkyBlue}100}}{6} \\ \\ & = 85.7\% \end{align}$ No the student will not receive the scholarship.

Assuming that the student gets a 90% final mark in their sixth class, which of the following will give the student the highest possible final average? Solution

Drop their lowest mark

Take an extra class, and get a 90% in it

Do a capstone project on any of the six classes to add 10% to the final mark

Do two capstone projects on any two of the six classes to add 5% each of the two final marks

If dropped lowest mark, $\begin{align} average & = \dfrac{88 + 83 + 85 + 86 + 90}{5} \\ \\ & = 86.4\% \end{align}$ Take an extra class, $\begin{align} average & = \dfrac{88 + 72 + 83 + 85 + 86 + 90 + 90}{7} \\ \\ & = 84.9\% \end{align}$ Do one capstone, $\begin{align} average & = \dfrac{88 + (72 + 10) + 83 + 85 + 86 + 90}{6} \\ \\ & = 85.7\% \end{align}$ Do two capstones, $\begin{align} average & = \dfrac{88 + (72 + 5) + (83 + 5) + 85 + 86 + 90}{6} \\ \\ & = 85.7\% \end{align}$ Therefore, dropping the lowest mark will give the student the highest possible final average.

If the student does a capstone project in any of the five classes to add 10% to the final mark, does it matter what class the student does the capstone project in? Solution

Yes

No

It shouldn't matter what class the student gets the extra percent in, because the sum and number of values will not differ. For each case the resulting calculation will be: $\begin{align} average & = \dfrac{sum}{number\ of\ values} \\ \\ & = \dfrac{414 + 10}{5} \\ \\ & = 84.8\% \end{align}$

Fraction	Percent
$\dfrac{2}{4}$	50 %
$\dfrac{3}{4}$	75 %
$\dfrac{2}{5}$
$\dfrac{1}{10}$	10%

Length (cm)	Width (cm)	Area (cm²)
1	___	___
2	___	___
3	___	___
5	___	___
6	___	___
9	___	___
10	___	___
15	___	___
18	___	___
30	___	___
45	___	___

Number of White Roses	Cost of Flower Arrangement
1	$3.00
4	$12.00
8	$24.00
12	$36.00
16	$48.00
24	$65.00

x	y
-2	-5(-2) + 3 = 10 + 3 = 13
-1	-5(-1) + 3 = 5 + 3 = 8
0	-5(0) + 3 = 0 + 3 = 3
1	-5(1) + 3 = -5 + 3 = -2
2	-5(2) + 3 = -10 + 3 = -7

Homeroom Classroom	Pepperoni	BBQ Chicken	No Preference
7A	10	12	8
8D	11	15	5
9F	9	9	9
9C	6	8	10
10G	13	13	0
10B	14	15	1
11T	21	1	3
11P	6	16	6
12G	4	8	14
12D	3	11	18

FAVORITES

OPTIONS

CLASS OPTIONS

GRADE

8

Math

TABLE OF CONTENTS

Number Sense, Numeration, Rates, Ratios

Representations of Powers: Pre-Product Rule

Simplify leaving your answer in the most reduced exponent form.

2 × 2 × 2 Solution ⁰¹²³⁴⁵⁶⁷⁸⁹⁻⁺⁽⁾₀₁₂₃₄₅₆₇₈₉₋₊₍₎ Incorrect Attempts: CHECK Hint Unavailable = 23

3 × 3 × 3 × 3 × 3 Solution ⁰¹²³⁴⁵⁶⁷⁸⁹⁻⁺⁽⁾₀₁₂₃₄₅₆₇₈₉₋₊₍₎ Incorrect Attempts: CHECK Hint Unavailable = 35

b × b × b Solution ⁰¹²³⁴⁵⁶⁷⁸⁹⁻⁺⁽⁾₀₁₂₃₄₅₆₇₈₉₋₊₍₎ Incorrect Attempts: CHECK Hint Unavailable = b3

c × c × c × c × c Solution ⁰¹²³⁴⁵⁶⁷⁸⁹⁻⁺⁽⁾₀₁₂₃₄₅₆₇₈₉₋₊₍₎ Incorrect Attempts: CHECK Hint Unavailable = c5

Scientific Notation: Expanded Form with Powers of × 10n

Convert to scientific notation.

148000 Solution Hint Clear Info × 10 Incorrect Attempts: CHECK Hint Unavailable

0.00000512 Solution Hint Clear Info × 10 Incorrect Attempts: CHECK Hint Unavailable

Understanding Powers of Ten with Scientific Notation

Convert the following scientific notation numbers.

1.725 × 10-9 Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable

(1.2 × 102) + (3.4 × 103) Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable

Comparing Fractions

Compare the following by filling in the blank (without using a calculator).

\quad \dfrac{16}{25} \ \text{____} \ \dfrac{7}{10} Solution Video > < = None of the above \dfrac{16}{25} \ < \ \dfrac{7}{10}

Converting Decimals and Fractions

Convert 0.75 to a fraction (remember to reduce fully). Solution \dfrac{7}{5} \dfrac{3}{4} \dfrac{75}{10} \dfrac{3}{2} \dfrac{3}{4}

Converting Decimals and Fractions

Convert \dfrac{250}{100} to a decimal. Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable 2.5

Factors and Multiples

The following numbers are factors of 22 Solution Video 22, 44, 66, 88, 110 True False These are multiples of 22. Multiples multiply, going up higher each time... There are infinite amount of multiples... (There are only a certain amount of factors... for 22 it's: 1, 2, 11, 22)

Determine the missing factor of 36: Solution 1, 2, 3, 4, 6, 9, 12, 18 Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable 36 is a factor of 36 because you can divide 36 by 36.

Fraction and Percent Word Problems

Jon wins $100. He gives 50% to his mother, and from what is left over after giving money to his mother, he gives \dfrac{2}{5} to charity. How much does Jon give to charity? Solution $ Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable 20 [steps not shown here]

Converting Decimals, Fractions, and Percents

Fill in the blank.

Solution FractionPercent \dfrac{2}{4}50 % \dfrac{3}{4}75 % \dfrac{2}{5} \dfrac{1}{10}10% Hint Clear Info Incorrect Attempts: CHECK % Hint Unavailable FractionPercent \dfrac{2}{4}50 % \dfrac{3}{4}75 % \dfrac{2}{5}40 % \dfrac{1}{10}10%

Solution DecimalPercent 0.330% 0.2525% 5% 0.011% Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable DecimalPercent 0.330% 0.2525% 0.055% 0.011%

Word Problems with Percents with One Decimal or Fewer

Determine the price or percent below, using a calculator.

13.5% of $49.99 Solution $ Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable \begin{align} final\ amount & = \dfrac{percent}{100} \times \ initial\ amount \\ \\ & = \dfrac{13.5\%}{100\%} \times \ $49.99 \\ \\ & = 0.135 \times \ $49.99 \\ \\ & = $6.75 \end{align}

Word Problems with Percents and Rates

The equation for earnings (from hourly rates and commissions) is given below. Show your work (in your own notes) and use a calculator to solve.

Adding, Subtracting, Multiplying, and Dividing Fractions

Simplify the following fraction expressions without the use of a calculator. Reduce your answer fully.

More Fractions with Order of Operations (PEMDAS/BEDMAS)

Simplify the following fraction expressions without the use of a calculator. Reduce your answer fully.

Order of Operations (PEMDAS, BEDMAS)

Simplify the following, without the use of a calculator.

Solution 6 x (4 ÷ 2) + 8 + (10 + 4) – 2 × 6 Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Use PEMDAS/BEDMAS...

Solution 3 × 4 ÷ 3 × 4 × (64 ÷ 16) Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Use PEMDAS/BEDMAS...

Solution Video (4 × 2) + 32 - 4 ÷ 2 Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Use PEMDAS/BEDMAS... \begin{align} & = (4 × 2) + 3^2 - 4 ÷ 2 \\ \\ & = (8) + 3^2 - 4 ÷ 2 \\ \\ & = 8 + 9 - 4 ÷ 2 \\ \\ & = 8 + 9 - 2 \\ \\ & = 15 \\ \\ \end{align}

Multiply and Divide by Powers of Ten

Solve, without the use of a calculator.

25 × 104 Solution Hint Clear Info Incorrect Attempts: CHECK Append 4 zeros to the end You start with 25.0 and add 4 zeros to the end (move the decimal 4 places, right).

123456 ÷ 103 = Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable You start with 123456.0 and move the decimal 3 places, left, to make the number smaller.

200 ÷ 10-5 Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable This is trick because the combination of '÷' with a negative exponents, is like a double negative, in that you end up moving the decimal place to the right. It's equivalent to 200 × 105.

Squares and Roots

Simplify

\sqrt{1^2} Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable A square root cancels out the 'perfect square' number. \begin{align} & = \sqrt{1^2} \\ \\ & = 1 \\ \\ \end{align}

\sqrt{3^2} Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable \begin{align} & = \sqrt{3^2} \\ \\ & = 3 \\ \\ \end{align}

\sqrt{4}\sqrt{81} Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable \begin{align} & = (2)(9) \\ \\ & = 18 \\ \\ \end{align}

2\sqrt{5^2}\sqrt{15} Solution Hint Clear Info \sqrt{\begin{matrix} \\ \quad \end{matrix}} Incorrect Attempts: CHECK Hint Unavailable \begin{align} & = 2\sqrt{5^2}\sqrt{15} \\ \\ & = 2(5)\sqrt{15} \\ \\ & = 10\sqrt{15} \\ \\ \end{align}

Proportional Relationships

Solve the following real-life proportion problems. You can use a calculator, if needed.

Equivalent Fractions as Proportional Relationships

Solve, without reducing.

Determine the answer. Hint Clear Info \dfrac{3}{4} \ × \ \dfrac{2}{2} = \dfrac{}{\quad\quad} Incorrect Attempts: CHECK Hint Unavailable \begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}

Determine parts of each fraction. Hint Clear Info \dfrac{5}{6} \ × \ \dfrac{3}{\quad\quad} = \dfrac{\quad\quad}{18} Incorrect Attempts: CHECK Hint Unavailable \begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}

Determine the first fraction. Hint Clear Info \dfrac{1}{3} \ × \ \dfrac{}{\quad\quad} = \dfrac{7}{21} Incorrect Attempts: CHECK Hint Unavailable \begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}

Measurement

Conversions Between m2 and cm2

Given the rectangle,

Calculate the area of the rectangle, in square centimeters. Solution Hint Clear Info Incorrect Attempts: CHECK cm2 Hint Unavailable Area = Length × Width \begin{align} & = 20 \ cm × 10 \ cm \\ \\ & = 200\ cm^2 \\ \\ \end{align}

Conversions of Square and Cubic Units

Convert the following, given,

2.5 m3 into cm3 Solution Hint Clear Info Incorrect Attempts: CHECK cm3 Hint Unavailable \begin{align} & = 2.5\ m^3 \dfrac{1,000,000\ cm^3}{1\ m^3} \\ \\ & = 2,500,000\ cm^3 \end{align}

663 cm3 into m3 Solution Hint Clear Info Incorrect Attempts: CHECK m3 Hint Unavailable \begin{align} & = 663\ cm^3 \dfrac{1\ m^3}{1,000,000\ cm^3} \\ \\ & = 0.000663\ m^3 \end{align}

Perimeter and Area (Circles)

The area of a circle is calculated with the equation, Area = πr2

2 × 2 × 2 Solution

⁰
¹
²
³
⁴
⁵
⁶
⁷
⁸
⁹
⁻
⁺
⁽
⁾
₀
₁
₂
₃
₄
₅
₆
₇
₈
₉
₋
₊
₍
₎

Incorrect Attempts:

CHECK

Hint Unavailable

= 2³

3 × 3 × 3 × 3 × 3 Solution

⁰
¹
²
³
⁴
⁵
⁶
⁷
⁸
⁹
⁻
⁺
⁽
⁾
₀
₁
₂
₃
₄
₅
₆
₇
₈
₉
₋
₊
₍
₎

Incorrect Attempts:

CHECK

Hint Unavailable

= 3⁵

b × b × b Solution

⁰
¹
²
³
⁴
⁵
⁶
⁷
⁸
⁹
⁻
⁺
⁽
⁾
₀
₁
₂
₃
₄
₅
₆
₇
₈
₉
₋
₊
₍
₎

Incorrect Attempts:

CHECK

Hint Unavailable

= b³

c × c × c × c × c Solution

⁰
¹
²
³
⁴
⁵
⁶
⁷
⁸
⁹
⁻
⁺
⁽
⁾
₀
₁
₂
₃
₄
₅
₆
₇
₈
₉
₋
₊
₍
₎

Incorrect Attempts:

CHECK

Hint Unavailable

= c⁵

Scientific Notation: Expanded Form with Powers of × 10ⁿ

148000 Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

Hint Unavailable

0.00000512 Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

Hint Unavailable

1.725 × 10^-9

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

(1.2 × 10²) + (3.4 × 10³)

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

$\quad \dfrac{16}{25} \ \text{____} \ \dfrac{7}{10}$ Solution Video

>

<

=

None of the above

$\dfrac{16}{25} \ < \ \dfrac{7}{10}$

Convert 0.75 to a fraction (remember to reduce fully). Solution

$\dfrac{7}{5}$

$\dfrac{3}{4}$

$\dfrac{75}{10}$

$\dfrac{3}{2}$

$\dfrac{3}{4}$

Convert $\dfrac{250}{100}$ to a decimal. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

2.5

The following numbers are factors of 22 Solution Video 22, 44, 66, 88, 110

True

False

These are multiples of 22. Multiples multiply, going up higher each time... There are infinite amount of multiples...

(There are only a certain amount of factors... for 22 it's: 1, 2, 11, 22)

Determine the missing factor of 36: Solution 1, 2, 3, 4, 6, 9, 12, 18

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

36 is a factor of 36 because you can divide 36 by 36.

Jon wins $100. He gives 50% to his mother, and from what is left over after giving money to his mother, he gives $\dfrac{2}{5}$ to charity. How much does Jon give to charity? Solution

$

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20 [steps not shown here]

Solution

Fraction Percent

$\dfrac{2}{4}$ 50 %

$\dfrac{3}{4}$ 75 %

$\dfrac{2}{5}$

$\dfrac{1}{10}$ 10%

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%

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Fraction Percent

$\dfrac{2}{4}$ 50 %

$\dfrac{3}{4}$ 75 %

$\dfrac{2}{5}$ 40 %

$\dfrac{1}{10}$ 10%

Solution

Decimal Percent

0.3 30%

0.25 25%

5%

0.01 1%

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Decimal Percent

0.3 30%

0.25 25%

0.05 5%

0.01 1%

13.5% of $49.99 Solution

$

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$\begin{align} final\ amount & = \dfrac{percent}{100} \times \ initial\ amount \\ \\ & = \dfrac{13.5\%}{100\%} \times \ $49.99 \\ \\ & = 0.135 \times \ $49.99 \\ \\ & = $6.75 \end{align}$

Solution 6 x (4 ÷ 2) + 8 + (10 + 4) – 2 × 6

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Use PEMDAS/BEDMAS...

Solution 3 × 4 ÷ 3 × 4 × (64 ÷ 16)

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Use PEMDAS/BEDMAS...

Solution Video (4 × 2) + 3² - 4 ÷ 2

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Use PEMDAS/BEDMAS... $\begin{align} & = (4 × 2) + 3^2 - 4 ÷ 2 \\ \\ & = (8) + 3^2 - 4 ÷ 2 \\ \\ & = 8 + 9 - 4 ÷ 2 \\ \\ & = 8 + 9 - 2 \\ \\ & = 15 \\ \\ \end{align}$

25 × 10⁴ Solution

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Append 4 zeros to the end

You start with 25.0 and add 4 zeros to the end (move the decimal 4 places, right).

123456 ÷ 10³ =

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You start with 123456.0 and move the decimal 3 places, left, to make the number smaller.

200 ÷ 10^-5

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This is trick because the combination of '÷' with a negative exponents, is like a double negative, in that you end up moving the decimal place to the right. It's equivalent to 200 × 10⁵.

$\sqrt{1^2}$ Solution

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A square root cancels out the 'perfect square' number. $\begin{align} & = \sqrt{1^2} \\ \\ & = 1 \\ \\ \end{align}$

$\sqrt{3^2}$ Solution

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$\begin{align} & = \sqrt{3^2} \\ \\ & = 3 \\ \\ \end{align}$

$\sqrt{4}\sqrt{81}$ Solution

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$\begin{align} & = (2)(9) \\ \\ & = 18 \\ \\ \end{align}$

$2\sqrt{5^2}\sqrt{15}$ Solution

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$\sqrt{\begin{matrix} \\ \quad \end{matrix}}$

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$\begin{align} & = 2\sqrt{5^2}\sqrt{15} \\ \\ & = 2(5)\sqrt{15} \\ \\ & = 10\sqrt{15} \\ \\ \end{align}$

Determine the answer.

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$\dfrac{3}{4} \ × \ \dfrac{2}{2} = \dfrac{}{\quad\quad}$

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Determine parts of each fraction.

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$\dfrac{5}{6} \ × \ \dfrac{3}{\quad\quad} = \dfrac{\quad\quad}{18}$

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Determine the first fraction.

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$\dfrac{1}{3} \ × \ \dfrac{}{\quad\quad} = \dfrac{7}{21}$

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$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Conversions Between m² and cm²

Calculate the area of the rectangle, in square centimeters. Solution

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cm²

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Area = Length × Width $\begin{align} & = 20 \ cm × 10 \ cm \\ \\ & = 200\ cm^2 \\ \\ \end{align}$

2.5 m³ into cm³ Solution

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cm³

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$\begin{align} & = 2.5\ m^3 \dfrac{1,000,000\ cm^3}{1\ m^3} \\ \\ & = 2,500,000\ cm^3 \end{align}$

663 cm³ into m³ Solution

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m³

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$\begin{align} & = 663\ cm^3 \dfrac{1\ m^3}{1,000,000\ cm^3} \\ \\ & = 0.000663\ m^3 \end{align}$

The area of a circle is calculated with the equation, Area = πr²

Which of the following can be used to calculate the amount of material required to make a circular mirror.

Radius

Circumference

Diameter

Any of the above

$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Determine the area of a circle with a diameter of 5 cm. Solution

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cm²

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Radius = Diameter ÷ 2 $\begin{align} Area & = πr^2 \\ \\ & = π(2.5)^2 \\ \\ & = 19.6\ cm^2 \end{align}$