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Tpy6a [65]
3 years ago
10

Thank you for your help :) really appreciate! Here’s the new one

Mathematics
2 answers:
Licemer1 [7]3 years ago
8 0

Answer

8 units

Explanation

The side of one of the squares is the same length as one of the legs on the triangle.

To find one side of the square, you need to find the square root, because to find the area of a square, you need to square the sides. To find one side, you do the opposite.

So the length of one side is the square root of 32.

Now don't worry that this is a weird number (which is 5.656854249...) because we'll be squaring it again.

The length of one side of the square is the leg of a right triangle.

To find the hypotenuse (or the longest side), you need to use the Pythagorean theorem.

l^{2}  + l^{2} = h^{2}

The square root of 32 squared is 32.

32+32=h^{2}

64=h^{2}

To find h, you need to find the square root. The square root of 64 is 8.

Therefore, the length of x is 8 units.

drek231 [11]3 years ago
4 0

Answer:

8 units

Step-by-step explanation:

The two squares have equal areas.

For each square:

area = 32 units^2

For a square:

area = side^2

side^2 = 32

side = sqrt(32)

For the two squares, the length of the side is sqrt(32).

The sides of the squares are the legs of the right triangle.

a^2 + b^2 = c^2

(sqrt(32))^2 + (sqrt(32))^2 = x^2

32 + 32 = x^2

x^2 = 64

x = 8

Answer: 8 units

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Answer:

54.5%

Step-by-step explanation:

Given data

Cost price= $15

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3 years ago
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Answer:

166.5 m³

Step-by-step explanation:

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If the parallel sides of a trapezoid are contained by the lines and , find the equation of the line that contains the midsegment
Fittoniya [83]

Answer:

Equation of midsegment line: y = (-1/4)x + 2.

Step-by-step explanation:

If the parallel sides of a trapezoid are contained by the lines:-

y = (-1/4)x +5 and y = (-1/4)x - 1

Midsegment of any trapezoid is the line segment

1. that is parallel to pair of parallel side of trapezoid and

2. that passes through the middle of the trapezoid and cuts the other two sides into equal-half.

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