The formula for the area of a trapezoid is A = a+b/2 * h. So just plug in the dimensions.
A = 5 + 3/2 * 5
A = 8/2 * 5
A = 4 * 5
A = 20 units²
I hope this helps love! :)

8 0

3 years ago

Help me with this...

marin [14]

i. 171

ii. 162

iii. 297

Solution,

n(U)= 630

n(I)= 333

n(T)= 168

i. Let n(I intersection T ) be X

$333 - x + x + 468 - x = 630 \\ or \: 333 + 468 - x = 630 \\ or \: 801 - x = 630 \\ or \: - x = 630 - 801 \\ or \: - x = - 171 \\ x = 171$

<h3>ii.n(only I)= n(I) - n(I intersection T)</h3><h3> = 333 - 171</h3><h3> = 162</h3>

<h3>iii. n ( only T)= n( T) - n( I intersection T)</h3><h3> = 468 - 171</h3><h3> = 297</h3>

<h3>Venn- diagram is shown in the attached picture.</h3>

Hope this helps...

Good luck on your assignment...

4 0

3 years ago

What’s the volume of the rectangular prism of 2cm, 7/3 cm and 2cm?

Eduardwww [97]

Answer:

9 1/3 cm³

Step-by-step explanation:

First multiply two numbers.

2x2=4

Then multiply the product by the third number.

4x7/3

4 x 7

1 x 3

= 28/3 or 9 1/3

hope it helps

3 0

3 years ago

Solve the system of linear equations by graphing. y = –5/2 x – 7 x + 2y = 4 What is the solution to the system of linear equatio

Anestetic [448]

Keywords

system; linear equations; graphing; solution

we have

$y=-\frac{5}{2}x-7$ -------> equation A

$x+2y=4$

$x=4-2y$ ----> equation B

The <u>system</u> of <u>linear equations</u> is composed of equation A and equation B

Substitute the equation B in equation A

$y=-\frac{5}{2}[4-2y]-7$

$y=-10+5y-7$

$4y=17$

$y=17/4=4.25$

Find the value of x

$x=4-2[17/4]=-4.5$

The<u> solution</u> is the point $(-4.5,4.25)$

Using a <u>graphing</u> tool

see the attached figure

therefore

the answer is the option A

$(-4.5,4.25)$

4 0

3 years ago

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