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4 years ago

What’s the area of the trapezoid

Mathematics

2 answers:

pickupchik [31]4 years ago

6 0

<h3>- - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - </h3>

➷ To work it out, you need to know this formula:

area of trapezoid = $\frac{a+b}{2} * h$

'a' is the top side

'b' is the bottom side

'h' is the vertical height

Substitute the values in:

area = $\frac{8 + 14}{2} * 12$

Now solve:

area = $132in^{2}$

In short, your answer is B.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ Hannah ♡

Aloiza [94]4 years ago

3 0

The answer is B. 132 in2

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Pls help me taking. Test I need answer fast

pychu [463]

Answer:

its b

Step-by-step explanation:

5 0

2 years ago

Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible

Law Incorporation [45]

Step-by-step explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;

$pq = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

$or \: \sqrt{( {5 + 3)}^{2} + ( {2 - 2)}^{2} }$

By simplifying it we get,

$8$

Now,

again finding the distance between PR,

$pr = \sqrt{( {x2 - x1}^{2} + ( {y2 - y1)}^{2} }$

or,

$\sqrt{( {x + 3)}^{2} + ( {y - 2)}^{2} }$

By simplifying it we get,

$= \sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 }$

now, finding the distance of QR,

$qr = \sqrt{( {x - 5)}^{2} + ( {y - 2)}^{2} }$

or, by simplification we get,

$\sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

now, equating PR and QR,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13} = \sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

we cancelled the root ,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = {x}^{2} + {y}^{2} -10x - 4y + 29$

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 } = 8}$

cancel the roots,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = 8$

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

x= 1

y=(8,-12)

So, i think it's (1,8) is your answer..

Hope it helps...

3 0

4 years ago

The product of 2 and a number minus 9 is 13

algol13

<h3>The value of the number is equal to 11.</h3>

2x - 9 = 13

Add 9 to both sides.

2x = 22

Divide both sides by 2.

x = 11

5 0

3 years ago

Solve the equation 1/4 (x+2)+5=-x

AnnyKZ [126]

Answer:

Exact Form:

x=−225

Decimal Form:

x=−4.4

Mixed Number Form:

x=−4 2/5

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%202x%2B31%3D5%7D" id="TexFormula1" title="\boxed{\sf 2x+31=5}" alt="\boxed{\s

AleksAgata [21]

2x + 31 = 5

2x = 5 - 31

2x = -26

x = -26/2

x = -13.

_______

꧁✿ ᴿᴬᴵᴺᴮᴼᵂˢᴬᴸᵀ2222 ✬꧂

3 0

3 years ago

Read 2 more answers