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mrs_skeptik [129]
2 years ago
13

If one diagonal of a rhombus is 15 meters long and it’s area is 157.5 square meters, find the measures of the other diagonal

Mathematics
1 answer:
ycow [4]2 years ago
7 0

Answer:

The other diagonal measures 21m

Step-by-step explanation:

In this question, we are tasked with calculating the length of the second diagonal of as Rhombus given the measure of the surface area of the rhombus and the length of the other diagonal

Mathematically, for a rhombus having two diagonals d_{1} and d_{2} , the area of the rhombus can be calculated mathematically using the formula below;

A = 1/2 ×d_{1} × d_{2}

From the question, we can identify that A = 157.5m^{2} and d_{1} = 15m

we input these in the formula;

157.5 = 1/2 × 15 × d_{2}

315 = 15 d_{2}

d_{2} = 315/15

d_{2} = 21m

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A defines the function

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f(3) = - 8 ×( \frac{5}{2}) ^{2} = - 8 × \frac{25}{4} = - 50

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Dylan is evaluating the expression 13 + 19 + 7 + 10. At one step in his work, Dylan rewrites the equation as 13 + 7 + 19 + 10. W
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Communicative Property

Step-by-step explanation:

The communicative property is basically just switching numbers around but still leading to the same answer, if you could see, both of the equations lead to the same answer, 49. Here is a chart to help you out.

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An economy package of cups has 460 green cups. If the green cups are 20% of the total package, how many cups are in the package?
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The formula for finding the perimeter of a rectangle is P = 2L + 2W. If a rectangle has a perimeter of 68 inches and the length
Artemon [7]

Answer:

Width = 10 inches

Step-by-step explanation:

Given the perimeter of a rectangle of 68 inches, and a length that is 14 inches longer than its width.

We can establish the following values to help us solve for the width of a rectangle:

Perimeter (P) = 68 inches

Length (L) = 14 + W inches

Width (W) = unknown

<h3 /><h3><u>Solve for the Width (W)</u></h3>

P =  2(L + W)  ⇒ This is the same as P = 2L + 2W, except that 2 is factored out from the right-hand side.

Divide both sides by 2:

\displaystyle\mathsf{\frac{P}{2}\:=\:\frac{2(L\:+\:W)}{2}}

\displaystyle\mathsf{\frac{P}{2}\:=L\:+\:W}

Substitute the value of the Perimeter and the length (L) into the formula:

\displaystyle\mathsf{\frac{68}{2}\:=14\:+W\:+\:W}

Combine like terms on the right-hand side, and simplify the left-hand side of the equation:

\displaystyle\mathsf{34\:=14\:+2W}

Subtract 14 from both sides:

34 - 14 = 14 - 14 + 2W

20 = 2W

Divide both sides by 2 to solve for the width (W):

\displaystyle\mathsf{\frac{20}{2}\:=\:\frac{2W}{2}}

W = 10 inches

Therefore, the width of the rectangle is 10 inches.

<h3 /><h3><u>Double-check:</u></h3>

Verify whether the derived value for the width is correct:

P = 2L + 2W

68 = 2(14 + 10) + 2(10)

68 = 2(34) + 20

68 = 48 + 20

68 = 68 (True statement).  

Thus, the length of the rectangle is 34 inches, and the width is 10 inches.

4 0
2 years ago
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