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blsea [12.9K]
3 years ago
5

What is the length of the hypotenuse of a right triangle that has sides measuring 12 cm and 35 cm

Mathematics
2 answers:
xz_007 [3.2K]3 years ago
8 0
37 is the answer becaus e12 squared plus 35 squared equals 1369 squared so that is 37
Westkost [7]3 years ago
3 0
The measurement is 37 cm
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EASY URGENT MATH EXERCISE ​
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Answer:

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Step-by-step explanation:

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the length of a rectangular sing is 5 time its width. if the sign perimeter is 24 inches. what is the sign area ?
kumpel [21]

Answer:

20 in²

Step-by-step explanation:

Assuming that the width of the sign is x, then from the question, we're told that the length is 5 times it's width, so

b = x inches

l = 5x inches

Again, we're told that the perimeter of the sign is 24 inches, and we know already that the perimeter of a rectangle is given as

Perimeter = 2(l + b), substituting this, we have

24 = 2(5x + x)

24 = 10x + 2x

24 = 12x

x = 24 / 12

x = 2 inches.

Since x is the width of the rectangle, and it's 2 inches, we use it to find the length of the sign.

l = 5 * 2

l = 10 inches.

Then, we are asked to find the area of the sign. Area of a rectangle is given as

A = l * b, so if we substitute, we have

Area = 10 * 2

Area = 20 square inches

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3 years ago
Fifteen students were asked which school day was least favorite. How many students chose Monday?
Lynna [10]
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3 years ago
Mr. Jones' other friend Ms. O'Neill has "x" meters of fencing, what is the maximum area in square meters she can have?
Alexus [3.1K]

Answer:

Max\ Area = \frac{x^2}{16}

Step-by-step explanation:

Given

P = x ---- the perimeter of fencing

Required

The maximum area

Let

L \to Length

W \to Width

So, we have:

P = 2(L + W)

This gives:

2(L + W) = x

Divide by 2

L + W = \frac{x}{2}

Make L the subject

L = \frac{x}{2} - W

The area (A) of the fence is:

A = L * W

Substitute L = \frac{x}{2} - W

A = (\frac{x}{2} - W) * W

Open bracket

A = \frac{x}{2}W - W^2

Differentiate with respect to W

A' = \frac{x}{2} - 2W

Set to 0

\frac{x}{2} - 2W = 0

Solve for 2W

2W = \frac{x}{2}

Solve for W

W = \frac{x}{4}

Recall that:

L = \frac{x}{2} - W

L = \frac{x}{2} - \frac{x}{4}

L = \frac{2x- x}{4}

L = \frac{x}{4}

So, the maximum area is:

A = L * W

A = \frac{x}{4}*\frac{x}{4}

Max\ Area = \frac{x^2}{16}

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3 years ago
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