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Oksi-84 [34.3K]
3 years ago
5

What two numbers if multiplied equal 144.585?

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
6 0
48.195*3
idk if u wanted any numbers that work but ya hope that helped :)
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(please explain clearly, I am very tired) Find the surface area of the composite figure.
Y_Kistochka [10]

Answer:

644

Step-by-step explanation:

SA=[2×(5×20 + 5×6 + 20×6)] +

[2×(4×12 + 4×6 + 12×6)] -

[ 2× 12×6]

= [2×(250)]+[2×(144)] - [144]

= 500+288-144

= 644 cm²

8 0
3 years ago
3. Find the smallest number of sweets that
White raven [17]

Answer:

9 = 3^2

15 = 3 x 5

20 = 2^2 x 5

24 =2^3 x 3

A = 2^3 x 3^2 x 5 =8 x 9 x 5 = 360

7 0
3 years ago
How would I solve <img src="https://tex.z-dn.net/?f=8%5E%7B3%2F2%7D%20%2A%202%5E%7B3%2F2%7D" id="TexFormula1" title="8^{3/2} * 2
andriy [413]
The asnwer is 36 i think

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6 0
3 years ago
Please help!!!!!!!!!!!!
Anna [14]

Answer:

5

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Answer only if you know<br><br> HELP PART A AND PART B
VARVARA [1.3K]

Answers:

  • Part A) There is one pair of parallel sides
  • Part B) (-3, -5/2) and (-1/2, 5/2)

====================================================

Explanation:

Part A

By definition, a trapezoid has exactly one pair of parallel sides. The other opposite sides aren't parallel. In this case, we'd need to prove that PQ is parallel to RS by seeing if the slopes are the same or not. Parallel lines have equal slopes.

------------------------

Part B

The midsegment has both endpoints as the midpoints of the non-parallel sides.

The midpoint of segment PS is found by adding the corresponding coordinates and dividing by 2.

x coord = (x1+x2)/2 = (-4+(-2))/2 = -6/2 = -3

y coord = (y1+y2)/2 = (-1+(-4))/2 = -5/2

The midpoint of segment PS is (-3, -5/2)

Repeat those steps to find the midpoint of QR

x coord = (x1+x2)/2 = (-2+1)/2 = -1/2

y coord = (x1+x2)/2 = (3+2)/2 = 5/2

The midpoint of QR is (-1/2, 5/2)

Join these midpoints up to form the midsegment. The midsegment is parallel to PQ and RS.

7 0
2 years ago
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