How do you solve #A = (1/2)h(b_1 + b_2)" for "b_2#?
Yes it is the formula for the Area A of the trapezoid
From the given formula
multiply both sides of the equation by 2
Divide both sides of the equation by h
by symmetric property
God bless....I hope the explanation is useful.
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To solve for ( b_2 ) in the equation ( A = \frac{1}{2}h(b_1 + b_2) ), first, multiply both sides by 2 to eliminate the fraction. Then, divide both sides by ( h ). Finally, subtract ( b_1 ) from both sides to isolate ( b_2 ). So, the solution is ( b_2 = \frac{2A}{h} - b_1 ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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