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LUCKY_DIMON [66]
3 years ago
15

What is the surface area of the square pyramid represented by the net?

Mathematics
2 answers:
AveGali [126]3 years ago
8 0
The area of the square is 36 because 6×6=36
The area of the triangles is 108 because (1/2)(6)(9)=27 27×4=108 because there are 4 triangles.

108+36=144
The correct answer is 144

Hope this helped :)

jonny [76]3 years ago
6 0

Answer:

144

Step-by-step explanation:

i go to k12

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(06.01) In the below system, solve for y in the first equation. x + 3y = 6 2x − y = 10 one thirdx + 2 negative one thirdx + 6 −x
NISA [10]

we have that

x + 3y = 6 -------> first equation

2x − y = 10


x + 3y = 6-----> substract x both sides

-x+x+3y=6-x

3y=6-x-----> divide by 3 both sides

y=(6-x) /3

y=2-(x/3)

y=-(x/3)+2

therefore


the answer is

negative one thirdx + 2

5 0
3 years ago
A van has room for 6students and 2teachers . How many vans are needed for a total of 48students and 16 teachers?
Juliette [100K]
8 vans are need for all to go
7 0
3 years ago
The table lists how poverty level income cutoffs​ (in dollars) for a family of four have changed over time. Use the midpoint for
frez [133]

The poverty level cutoff in 1987 to the nearest dollar  was $10787.

<h3>How to find a midpoint?</h3>

The midpoint as the point that divides the line segment exactly in half having two equal segments. Therefore, the midpoint presents the same distance between the endpoints for the line segment. The midpoint formula is: \mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right).

For solving this exercise, first you need plot the points in a chart. See the image.

Your question asks to approximate the poverty level cutoff in 1987 to the nearest dollar using the midpoint formula. Note that the year 1987 is between 1980 and 1990, thus you should apply the midpoint formula from data for this year (1987).

\mathrm{Midpoint\:of\:}\left(1980,8429),\:\left(1990,13145):\quad \left(\frac{1990+1980}{2},\:\:\frac{13145+8429}{2}\right)\\\\ \\

\mathrm{Midpoint\:of\:}\left(1980,8429)\:=\left(\frac{1990+1987}{2},\:\frac{13145+3843}{2}\right)\\ \\ =\left(\frac{3977}{2},\:\frac{21574}{2}\right)

\mathrm{Midpoint\:of\:}\left(1980,8429)\:=\=(\frac{3977}{2},10787)

The answer for your question will be the value that you calculated for the y-coordinate. Then, the poverty level cutoff in 1987 to the nearest dollar  was $10787.

Read more about the midpoint segment here:

brainly.com/question/11408596

7 0
2 years ago
Simplify 3(2x-2)-2(x-2)<br><br> A.3x-3<br> B.3x-4<br> C.4x-2<br> D.4x-4
Black_prince [1.1K]
The answer is c 4x-2
5 0
3 years ago
Read 2 more answers
23/(3+4)-12=??<br> help ;-;
aalyn [17]

Answer:

-4.6

because you add whats inside teh parenthesis then subtract by 12 before dividing by 23

8 0
2 years ago
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