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jarptica [38.1K]
3 years ago
7

If you don’t know the real answer don’t give me a fake one I just need help!

Mathematics
1 answer:
PtichkaEL [24]3 years ago
8 0
Answer:
1. x + 24 = 3x
24 = 3x - x
24 = 2x
12 = x

3x
3 (12)
36

2. 10x - 27 = 7x
10x - 7x = 27
3x = 27
x = 9

7x
7 (9)
63

3. 5x + 1 = 6x - 13
5x - 6x = -13 - 1
-x = -14
x = 14

5x + 1
5 (14) + 1
71

4. 6x + 3 = 4x + 39
6x - 4x = 39 - 3
2x = 36
x = 18

6x + 3
6 (18) + 3
111

1. B
2. A
3. H
4. E

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Tpy6a [65]
Draw a rectangle and label the right side as L-3 and the bottom side as L.

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Look at the figure XYZ in the coordinate plane. Find the perimeter of the figure rounded to the nearest tenth.
ki77a [65]

the distance form X to Y is clearly -6 to 0 is 6 units, and 0 to 8 is 8 units, so 6 + 8 = 14 units.

now, for XZ and ZY we can simply use as stated, the distance formula to get those and then add them all to get the perimeter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ X(\stackrel{x_1}{-6}~,~\stackrel{y_1}{2})\qquad Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ XZ=\sqrt{[5-(-6)]^2+[8-2]^2}\implies XZ=\sqrt{(5+6)^2+(8-2)^2} \\\\\\ XZ=\sqrt{121+36}\implies \boxed{XZ=\sqrt{157}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad Y(\stackrel{x_2}{8}~,~\stackrel{y_2}{2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ ZY=\sqrt{(8-5)^2+(2-8)^2}\implies ZY=\sqrt{9+36}\implies \boxed{ZY=\sqrt{45}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{perimeter}{14+\sqrt{157}+\sqrt{45}}\qquad \approx \qquad 33.2

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3 years ago
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