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4 years ago

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An isosceles triangle is a triangle in which two sides are equal in length. the angle between the two equal sides is called the

vertex angle, while the other two angles are called the base angles. if the vertex angle is 40°, what is the measure of the base angles?

1 answer:

Murrr4er [49]4 years ago

4 0

It's 180-40 divide 2 which equals = 70

I hope it helped you!!!

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Alecsey [184]

The equations have been computed below

<h3>How to illustrate the information?</h3>

3(x + 2) + 11 = 5(x - 4)

3x + 6 + 11 = 5x - 20

Collect like terms

3x - 5x = -20 - 17

-2x = -37

x = 37/2

x = 18.5

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6x + 4 = 9 - 6x

Collect like terms

6x + 6x = 9 - 4

12x = 5.

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2 years ago

an 8-inch by 1-inch rectangular design is being enlarged on a photocopier by a factor of 120%. what are the new dimensions of th

gizmo_the_mogwai [7]

9.6 inches by 1.2 inches

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4 years ago

In what year did the US stock market collapsed?

dexar [7]

The US stock market collapsed in 1929.

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4 years ago

What is equivalent too 4a + 10

34kurt

The answer is 2(2a+5)

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3 years ago

The length of triangle base is 36. The line which is parallel to the base divides the triangle into two equal area parts. Find a

Soloha48 [4]

Answer:

$18\sqrt{2}$

Step-by-step explanation:

Consider triangle ABC with the base AC=36. Let the line DE be parallel to the line AC. If DE||AC, then triangles ABC and DBE are similar. If k is the scale factor of these triangles, then

$\dfrac{DE}{AC}=\dfrac{BG}{BF}=k.$

Thus,

$DE=k\cdot AC=36k,\\ \\BG=k\cdot BF.$

The area of the triangle ABC is

$A_{ABC}=\dfrac{1}{2}AC\cdot BF=\dfrac{1}{2}\cdot 36\cdot BF=18BF.$

The area of the triangle DBE is

$A_{DBE}=\dfrac{1}{2}DE\cdot BG=\dfrac{1}{2}\cdot 36k\cdot kBF=18k^2BF.$

Since line DE divides triangle ABC in two equal area parts, we have that

$A_{DBE}=\dfrac{1}{2}A_{ABC},\\ \\18k^2BF=\dfrac{1}{2}\cdot 18BF,\\ \\k^2=\dfrac{1}{2},\\ \\k=\dfrac{1}{\sqrt{2}}$

and

$DE=\dfrac{1}{\sqrt{2}}\cdot 36=18\sqrt{2}.$

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3 years ago