The fact that we can evaluate an expression (or formula) when we know the values of the variables used forms the bedrock of Algebra. We will be using this constantly from here on out, both directly and indirectly, to make sense of all other concepts that we will see.
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1. Find the volume (V) of a cone using the formula $V=\pi r^2 \frac{h}{3}$
when the radius (r) is $5$ cm and the height (h) is $9$ cm. Use 3.14 for pi ($\pi$).
| Solution: | $V=\pi r^2 \frac{h}{3}$ |
|---|---|
| Substitute the given values into the formula | $V=3.14 * 5^2 * \frac{9}{3}$ |
| Use order of operations to simplify the expression | |
| Exponents | $V= 3.14 * 25 * \frac{9}{3}$ |
| Divide and Multiply | $V=78.5 * \frac{9}{3}$ |
| from left to right | $V=78.5 * 3$ |
| $V=235.5 cm^3$ |
2. Evaluate $(-1+r)^3$ when $r= \frac{1}{4}$. Write your answer as a fraction.
| Solution: | $(-1+r)^3$ |
|---|---|
| Substitute the given value into the expression. | $(-1+ \frac{1}{4})^3$ |
| Use order of operations to simplify the expression | $(-1+ \frac{1}{4})^3$ |
| Parenthese | $(- \frac{3}{4})^3$ |
| Exponent | $(- \frac{3}{4})^3 = – \frac{3}{4} * – \frac{3}{4} * – \frac{3}{4} =- \frac{9}{64}$ |
