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eimsori [14]
3 years ago
6

The following is the formula for finding the area (A) of a trapezoid with height (h) and bases (b1) and (b2). A = 1/2h(b1 + b2)

Which shows the formula correctly solve for the height in terms of the two bases?
A. h = 1/2A(b1 + b2)
B. h = 2A/b1 + b2
C. h = 2A - b1 - b2
D. h = A/2(b1 + b2)

Mathematics
1 answer:
charle [14.2K]3 years ago
6 0

Answer:

<em>Correct choice: B</em>

Step-by-step explanation:

<u>Equation Solving</u>

The area of a trapezoid with height h and bases b1 and b2 is given by:

\displaystyle A=\frac{1}{2}h(b_1+b_2)

We must solve this formula for h.

First, multiply by 2 to eliminate denominators:

\displaystyle 2A=h(b_1+b_2)

Now, divide by b1+b2:

\displaystyle \frac{2A}{b_1+b_2}=h

Swapping sides:

\displaystyle h=\frac{2A}{b_1+b_2}

Correct choice: B

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\huge \bf༆ Answer ༄

Let the capacity of bus be x students

And van be y students, now ;

From the given statements we get two equations ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260 \:   \:  \:  \: \:  (1)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:13x + 5y = 670 \:  \:  \:  \:  \: (2)

multiply the equation (2) with 2 [ it won't change the values ]

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260 \: \:  \:  \:   \:  \:  \:  \:  \: (1)

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Now, deduct equation (1) from equation (3)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:26x + 10y - 2x - 10y = 1340 - 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:24x = 1080

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x = 1080 \div 24

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x = 45

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Now, plug the value of x in equation (1) to find y ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:(2 \times 45) + 10y = 260

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{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:10y = 260 - 90

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{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 170 \div 10

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 17

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