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3 years ago

What percent of 88 is 72.6? ...?

1 answer:

8 0

Percentage problems are simple division problems. Here, what percentage of x is y means the answer is y divided by x (times 100). Let's apply it to this question.
Divide 72.6 by 88, 72.6/88. So the answer would be 0.825 then multiply it to 100 to make it a whole. The answer would 82.5%. I hope this helps you with your problem

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using the graphing method to solve the equation 13(0.25)^x=7 Round your answer to the nearest thousandth

netineya [11]

Answer:

x = 0.447

Step-by-step explanation:

You need to use your graphing calculator for this.

In y=, put 13(0.25)^x in Y1 and put 7 in y2. Using 2nd trace, find the intersections.

The answer I got was x = 0.447.

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3 years ago

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Lina20 [59]

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In physics, kinetic energy is the energy of an object in motion. Kinetic energy, K, measured in joules (J), is

Kryger [21]

Answer:

$m = \frac{2k}{v^{2} }$

Step-by-step explanation:

$k = \frac{1}{2} mv ^{2}$

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3 years ago

The radius of a circle is 14.2 m. Find the circumference to the nearest tenth.

lana66690 [7]

Answer:

C=89.2

Step-by-step explanation:

The formula of the circumference of a circle is C=2piR (or C=piD). Now, when calculating, plug in the numbers, giving you C=2pi14.2. Multiply everything, you get a super long irrational number; just round. You will get 89.2.

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2 years ago

Which of these four sets of side lengths will form a right triangle? Set 1 4 cm, 5 cm, 6 cm Set 2 8 in., 12 in., 20 in. Set 3 10

Alenkasestr [34]

Answer: The correct option is (D) SET 4.

Step-by-step explanation: We are to select the correct set of side lengths that will form a right-angled triangle.

To form a right-angled triangle, we must have the following relation:

Perpendicular² + Base² = Hypotenuse².

Hypotenuse is the length of the largest side; perpendicular and base are the two legs of the triangle.

SET 1 : 14 cm, 5 cm, 6 cm.

We have

$5^2+6^2=25+36=61,\\\\14^2=196.$

Therefore,

Perpendicular² + Base² ≠ Hypotenuse².

So, this set will not form a right-angled triangle.

SET 2 : 8 in., 12 in., 20 in.

We have

$8^2+12^2=64+144=208,\\\\20^2=400.$

Therefore,

Perpendicular² + Base² ≠ Hypotenuse².

So, this set will not form a right-angled triangle.

SET 3 : 10 mm, 20 mm, 30 mm.

We have

$10^2+20^2=100+400=500,\\\\30^2=900.$

Therefore,

Perpendicular² + Base² ≠ Hypotenuse².

So, this set will not form a right-angled triangle.

SET 4 : 12 ft, 16 ft, 20 ft.

We have

$12^2+16^2=144+256=400,\\\\20^2=400.$

Therefore,

Perpendicular² + Base² = Hypotenuse².

So, this set will form a right-angled triangle.

Thus, the SET 4 will form a right-angles triangle.

Option (D) is correct.

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3 years ago

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