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sertanlavr [38]
3 years ago
15

Solve A=h/2(b1 + b2) for h. Is this like the one before? We are a bit confused. ...?

Chemistry
2 answers:
Masja [62]3 years ago
8 0
A=(h)/(2)*(b+b^(2)) 

<span>Since h is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. </span>
<span>(h)/(2)*(b+b^(2))=a </span>

<span>Reorder the polynomial b+b^(2) alphabetically from left to right, starting with the highest order term. </span>
<span>(h)/(2)*(b^(2)+b)=a </span>

<span>Factor out the GCF of b from each term in the polynomial. </span>
<span>(h(b(b)+b(1)))/(2)=a </span>

<span>Factor out the GCF of b from b^(2)+b. </span>
<span>(h(b(b+1)))/(2)=a </span>

<span>Multiply h by b to get bh. </span>
<span>((bh)(b+1))/(2)=a </span>

<span>Remove the parentheses from the numerator. </span>
<span>(bh(b+1))/(2)=a </span>

<span>Multiply each term in the equation by 2. </span>
<span>(bh(b+1))/(2)*2=a*2 </span>

<span>Simplify the left-hand side of the equation by canceling the common terms. </span>
<span>bh(b+1)=a*2 </span>

<span>Multiply a by 2 to get 2a. </span>
<span>bh(b+1)=2a </span>

<span>Divide each term in the equation by (b+1). </span>
<span>(bh(b+1))/(b+1)=(2a)/(b+1) </span>

<span>Simplify the left-hand side of the equation by canceling the common terms. </span>
<span>bh=(2a)/(b+1) </span>

<span>Divide each term in the equation by b. </span>
<span>(bh)/(b)=(2a)/(b+1)/(b) </span>

<span>Remove the common factors that were cancelled out. </span>
<span>h=(2a)/(b+1)/(b) </span>

<span>Simplify the right-hand side of the equation by simplifying each term. </span>
<span>h=(2a)/(b(b+1))

--------------------------------------------------------

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
natulia [17]3 years ago
4 0
Do the total opposite operations until u get to H by itself

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