All categories

3 years ago

The area of the trapezoid is 108 in2. What is the height of the trapezoid?

Mathematics

2 answers:

arsen [322]3 years ago

6 0

Area for a trapezoid is A= a+b/2 times h (a and b are the bases which are 8 and 10

so 108=8+10/2 x h 8+10 divided by 2 equals 9 so now we have 108=9 x h

now we must divide both sides by 9 to solve for the height 108/9 equals 12 so the height of the trapezoid is 12

dezoksy [38]3 years ago

5 0

The area of the trapezoid is given by the formula A= (a+b)/ 2 * h

in this question a= 8, b= 10, A= 108 inches², and h is what we have to find, substitute what we have got into the formula and you will have:

108= (8+10) /2 *h

108= 9 *h

h= 108/9

h= 12 inches

You might be interested in

(graph photo attached)

Mnenie [13.5K]

B negative 0.9 is right

8 0

4 years ago

The rate of change for a quadratic function is

Tema [17]

Answer:

Step-by-step explanation:

hjhj

7 0

3 years ago

Geometry help 10 points

Serhud [2]

The answer to the reflection question would be D.

3 0

3 years ago

Read 2 more answers

Zoe's pizza cost $12.99. She gave a tip of $2.00.

Ksenya-84 [330]

I think it’s A I may be wrong tho

8 0

3 years ago

Read 2 more answers

Derivative of these questions<br>

shtirl [24]

Answer: $\bold{y'=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)}$

<u>Step-by-step explanation:</u>

$y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\\\\\\\text{Rationalize the denominator:}\\y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\bigg(\dfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}\bigg)\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{x+a-x-b}\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}\\\\\\\text{Apply the derivative (derivative of the numerator (top) divided by the}\\\text{denominator (bottom) squared).}$

$\text{Derivative of }\sqrt{x+a}\quad \rightarrow \quad (x+a)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + a)^{-\frac{1}{2}}\\\\\text{Derivative of }\sqrt{x+b}\quad \rightarrow \quad (x+b)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + b)^{-\frac{1}{2}}\\\\\\\text{Derivative of}\ \dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}:\\\\= \dfrac{\dfrac{1}{2}(x + a)^{-\frac{1}{2}}+\dfrac{1}{2}(x + b)^{-\frac{1}{2}}}{(a-b)^2}$

$\text{Factor out }\dfrac{1}{2}\ \text{from the numerator:}\\\\\dfrac{(x+a)^{-\frac{1}{2}}+(x+b)^{-\frac{1}{2}}}{2(a-b)^2}\\\\\\\text{Move the terms with negative exponents to the denominator:}\\\\=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)$

3 0

3 years ago