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))

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. 
(h)/(2)*(b+b^(2))=a 

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

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

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

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

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

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

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

Multiply a by 2 to get 2a. 
bh(b+1)=2a 

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

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

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

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

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

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