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1 year ago

In triangle ABC, m∠A = 36°, m∠B = 84°, and m∠C = 60°. The side lengths, in order from greatest to least, are , , .

Mathematics

1 answer:

katen-ka-za [31]1 year ago

5 0

In the triangle ABC, the side lengths, in order from the greatest to the least, are : AC > AB > BC.

We are given a triangle. The vertices of the triangle are A, B, and C. The measures of the angles ∠A, ∠B, and ∠C are 36°, 84°, and 60°, respectively. We need to arrange the side lengths in order from the greatest to the least.

The side lengths are proportional to their opposing angles in a triangle. It means that the side opposite the largest angle is the largest side, and vice versa. The angles arranged in descending order are : 84° > 60° > 36°. The angles arranged in descending order according to the vertices are : B > C > A. The order of the lengths of the opposite sides must be the same.

Hence, the side lengths, in order from the greatest to the least, are : AC > AB > BC.

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find the difference between the sum of all eight positive intergral divisors of 66 and the sum of all eight positive integral di

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The difference between the sum of all eight positive integral divisors of 66 and the sum of all eight positive integral divisors of 70 is zero.

<h3>How to find the difference between the integral divisors?</h3>

First let's find the integral divisors. We can write 66 as a product of prime numbers as:

66 = 33*2 = 2*3*11

Then the integral divisors of 66 are:

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1 (trivially)

66 (trivially)

The sum gives:

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1 year ago

Add 7 to me. Then subtract 10. If you subtract 5 and then add 9, you get 14. What number am I?

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Answer:

your answer is 13.

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3 years ago

Kite WXYZ is graphed on a coordinate plane. What is the approximate perimeter of the kite? Round to the nearest tenth. 10.6 unit

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<u>Answer-</u>

<em>Perimeter of the kite is </em><em>16.2 units</em>

<u>Solution-</u>

As WXYZ is a kite, so two disjoint pairs of consecutive sides are congruent, i.e WX=XY and WZ=ZY

So, perimeter of the kite WXYZ is,

$=2(\overline{WX}+\overline{WZ})$

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$\overline{WZ}=\sqrt{(1-3)^2+(1+3)^2}=\sqrt{(-2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}$

So, perimeter will be,

$P=2(\sqrt{13}+\sqrt{20})=16.15\approx 16.2\ units$

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