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4 years ago

Help with these 4 EQUATIONS . ( PLEASE SHOW WORK) WILL MARK AS BRAINLIEST

Mathematics

1 answer:

Kazeer [188]4 years ago

3 0

8x+y=18
Y=3x-22
8x+3x-22=18
11x=40
X=40/11
Y=3×40/11-22
Y=-122/11
X=40/11
Y=-122/11

X+8y=17
X=y+8
Y+8+8y=17
9y=9
Y=1
X=1+8
X=9
Y=1

9x+y=58 Y=2x-8 9x+2x-8=58 11x=66 X=6 Y=2(6)-8 Y=12-8 Y=4 X=6
X+8y=21 X=2y+1 2y+1+8y=21 10y=20 Y=2 X=2 (2)+1 X=5 Y=2

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In ∆PQR, p=13 inches, q=18 inches and r= 12 inches. Find the area of ∆PQR to the nearest square inch.

sveta [45]

Given data:

The first side of the triangle is p=13 inches.

The second side of the triangle is q=18 inches.

The third side of the triangle is r= 12 inches.

The semi-perimeter is,

$\begin{gathered} s=\frac{p+q+r}{2} \\ =\frac{13\text{ in+18 in+12 in}}{2} \\ =21.5\text{ in} \end{gathered}$

The expression for the area of the triangle is,

$\begin{gathered} A=\sqrt[]{s(s-p)(s-q)(s-r)_{}} \\ =\sqrt[]{21.5\text{ in(21.5 in-13 in)(21.5 in-18 in)(21.5 in-12 in)}} \\ =\sqrt[]{(21.5\text{ in)(8.5 in)(3.5 in)(9.5 in)}} \\ =77.95in^2 \end{gathered}$

Thus, the area of the given triangle is 77.95 sq-inches.

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