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Elden [556K]
4 years ago
7

Find the area of an equilateral triangle (regular 3-gon) with the given measurement.

Mathematics
2 answers:
notka56 [123]4 years ago
3 0
A= (1/2) b*h
h=3 sqrt3
A= (1/2)*6*3sqrt3
A= 9sqrt3 square inches
Alisiya [41]4 years ago
3 0

Answer:

The area of the triangle is 9√3 in²

Step-by-step explanation:

We know that, the area of an equilateral triangle is,

A=\frac{\sqrt{3}}{4}a^2

Where, a is the side of the triangle,

Given,

a = 6 inches,

Thus, the area of the given equilateral triangle is,

A=\frac{\sqrt{3}}{4}(6)^2

=\frac{36\sqrt{3}}{4}

=9\sqrt{3}\text{ square inches}

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Answer:

1) For each value of x, a value of y is increased by 5.

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2) For each two values of x, a value of y is increased by 10.

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Step-by-step explanation:

This is as easy as replacing x for the actual value show on the table.

1)

y=5x+5

When x = 0, y = ?

y=5(0)+5\\y=0+5\\y=5

When x = 1, y = ?

y=5(1)+5\\y=5+5\\y=10

When x = 2, y = ?

y=5(2)+5\\y=10+5\\y=15

When x = 3, y = ?

y=5(3)+5\\y=15+5\\y=20

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2)

y=5x-2

When x = 0, y = ?

y=5(0)-2\\y=0-2\\y=-2

When x = 2, y = ?

y=5(2)-2\\y=10-2\\y=8

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When x = 6, y = ?

y=5(6)-2\\y=30-2\\y=28

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3)

y=\frac{2}{5}x +1

When x = 0, y = ?

y=\frac{2}{5}(0) +1

y=0 +1\\y=1

When x = 1, y = ?

y=\frac{2}{5}(1) +1\\y=\frac{2}{5} +1

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y=\frac{2}{5}(5) +1\\y=2+1\\y=3

When x = 10, y = ?

y=\frac{2}{5}(10) +1\\y=2(2) +1\\y=4+1\\y=5

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When x = 5, y = ?

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