All categories

2 years ago

A trapezoid has an area of 50 square inches.

Mathematics

1 answer:

Shalnov [3]2 years ago

4 0

The area of a trapezoid can be computed as:

A=(b1+b2)h/2, where b1 and b2 ar the bases, h is the height.

With the provided numerical values:

50=(3+7)h/2

50=10h/2=5h

Answer: h=10inches

You might be interested in

One year, in school A, 162 students leave and 109 go to university. In school B, 75 students leave and 68 go to university. Comp

eduard

Answer:

in school a , 67.3 % of leaving students go to university.

in school b ,90.7% of leaving students go to university

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

X^2-2xy+y2 is a polynomial of ___________variable?

fgiga [73]

Answer:

x and y

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the true solution to 2 In ein 5x = 2 In 15

inessss [21]

Whyegevvdvevehuevdhehshh hsuwiuwhehehhehehev hehehehhehehevvshsisiehbebe uehehebbeheh hehehehhe. Hehehebvege. Hehehehevve be

6 0

2 years ago

Solve 732 H5 71z so what would x equal?

marin [14]

If the bases ar same, assume exponents are same

if
x^z=x^m
assume z=m

so
base is both 7 therefor
3x+5=1-x
solve
add x
4x+5=1
minus 5
4x=-4
divide 4
x=-1

7 0

3 years ago

Apply Crammer's Rule to find the solution to the following quations .2x + 3y = 1, 3x + y = 5

Bond [772]

Answer:

The solution to the equation system given is:

<u>x = 2</u>
<u>y = -1</u>

Step-by-step explanation:

First, we must know the equations given:

2x + 3y = 1
3x + y = 5

Following Crammer's Rule, we have the matrix form:

$\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] =\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]$

Now we solve using the determinants:

$x=\frac{\left[\begin{array}{ccc}1&3\\5&1\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(1*1)-(5*3)}{(2*1)-(3*3)} = \frac{1-15}{2-9} =\frac{-14}{-7} = 2$

$y=\frac{\left[\begin{array}{ccc}2&1\\3&5\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(2*5)-(3*1)}{(2*1)-(3*3)}=\frac{10-3}{2-9} =\frac{7}{-7}=-1$

Now, we can find the answer which is x= 2 and y= -1, we can replace these values in the equation to confirm the results are right, with the first equation:

2x + 3y = 1
2(2) + 3(-1)= 1
4 - 3 = 1
1 = 1

And, with the second equation:

3x + y = 5
3(2) + (-1) = 5
6 - 1 = 5
5 = 5

4 0

2 years ago