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Solve for h:A = ¹h (b₁ + b₂)Oh 2A - b₁ + b₂Oh==0 h =2Ab₁+ b₂4bi+by

Solve for h:A = ¹h (b₁ + b₂)Oh 2A - b₁ + b₂Oh==0 h =2Ab₁+ b₂4bi+by-example-1
User Rayimag
by
8.3k points

1 Answer

7 votes

Given:

There are given the formula:


A=(1)/(2)h(b_1+b_2)

Step-by-step explanation:

According to the question:

We need to find the value of h.

So,

First, multiply by 2 on both sides of the equation.

Then,


\begin{gathered} A=(1)/(2)h(b_(1)+b_(2)) \\ 2A=(1)/(2)h(b_1+b_2)*2 \\ 2A=h(b_1+b_2) \end{gathered}

Then,


\begin{gathered} 2A=h(b_(1)+b_(2)) \\ h=(2A)/(b_1+b_2) \end{gathered}

Therefore, the value of h is:


h=(2A)/(b_(1)+b_(2))

Final answer:

Hence, the correct option is B.

User Fsiaonma
by
8.2k points
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