Given $$ A=\frac{1}{2}\left(b_{1}+b_{2}\right) h $$, solve for $$ h $$ A $$ h=2 A\left(b_{1}+b_{2}\right) $$ B $$ \mathrm{h}=\frac{b_{1}+b_{2}}{2 A} $$ C $$ \mathrm{h}=\frac{2 A}{b_{1}+b_{2}} $$ D $$ h=A-\frac{1}{2}\left(b_{1}+b_{2}\right) $$

Question

UpStudy AI · Accepted Answer

To solve the equation $$ A = \frac{1}{2}(b_1 + b_2)h $$ for $$ h $$, follow these steps:

1. **Start with the original equation:**
   $$
   A = \frac{1}{2}(b_1 + b_2)h
   $$

2. **Multiply both sides by 2 to eliminate the fraction:**
   $$
   2A = (b_1 + b_2)h
   $$

3. **Divide both sides by $$ (b_1 + b_2) $$ to solve for $$ h $$:**
   $$
   h = \frac{2A}{b_1 + b_2}
   $$

Thus, the correct solution for $$ h $$ is:

**C. $$ \mathrm{h}=\frac{2 A}{b_{1}+b_{2}} $$**
$$ h = \frac{2A}{b_1 + b_2} $$, choice C.

Given , solve for

A

B

C

D

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Algebra Questions

Given , solve for A B C D