Question

a.ced.a.4: transforming formulas (expr
problem 11
21 when the formula $p = 2l + 2w$ is solved for $w$, the
result is

1. $w = \frac{2l + p}{2}$
2. $w = \frac{p - 2l}{2}$
3. $w = \frac{p}{2} + l$
4. $w = l - \frac{p}{2}$

problem 12
22 the formula for the area of a trapezoid is
$a = \frac{1}{2}h(b_1 + b_2)$. the height, $h$, of the trapezoid
may be expressed as

1. $\frac{2a}{b_1 + b_2}$
2. $\frac{1}{2}a(b_1 + b_2)$
3. $\frac{b_1 + b_2}{2a}$
4. $\frac{1}{2}a - (b_1 + b_2)$

Explanation:

Problem 11

Step1: Isolate term with $w$

$p - 2l = 2w$

Step2: Solve for $w$

$w = \frac{p - 2l}{2}$

Problem 12

Step1: Eliminate fraction coefficient

$2A = h(b_1 + b_2)$

Step2: Solve for $h$

$h = \frac{2A}{b_1 + b_2}$

Answer:

Problem 11: 2) $w = \frac{p - 2l}{2}$
Problem 12: 1) $\frac{2A}{b_1 + b_2}$