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Pie
2 years ago
6

Nequalities in Non Standard Form

Mathematics
1 answer:
Dominik [7]2 years ago
7 0

Answer:

x < 31 I think

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They both use the same units of measure.

Step-by-step explanation:

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The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?
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we have

A(-2, 2),B(6, 2),C(0, 8)

see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

P=AB+BC+AC

and

the area of the triangle is equal to

A=\frac{1}{2}*base *heigth

in this problem

base=AB\\heigth=DC

we know that

The distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Step 1

<u>Find the distance AB</u>

A(-2, 2),B(6, 2)

Substitute the values in the formula

d=\sqrt{(2-2)^{2}+(6+2)^{2}}

d=\sqrt{(0)^{2}+(8)^{2}}

dAB=8\ units

Step 2

<u>Find the distance BC</u>

B(6, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0-6)^{2}}

d=\sqrt{(6)^{2}+(-6)^{2}}

dBC=6\sqrt{2}\ units

Step 3

<u>Find the distance AC</u>

A(-2, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0+2)^{2}}

d=\sqrt{(6)^{2}+(2)^{2}}

dAC=2\sqrt{10}\ units

Step 4

<u>Find the distance DC</u>

D(0, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0-0)^{2}}

d=\sqrt{(6)^{2}+(0)^{2}}

dDC=6\ units

Step 5

<u>Find the perimeter of the triangle</u>

P=AB+BC+AC

substitute the values

P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units

P=22.81\ units

therefore

The perimeter of the triangle is equal to 22.81\ units

Step 6

<u>Find the area of the triangle</u>

A=\frac{1}{2}*base *heigth

in this problem

base=AB=8\ units\\heigth=DC=6\ units

substitute the values

A=\frac{1}{2}*8*6

A=24\ units^{2}

therefore

the area of the triangle is 24\ units^{2}

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