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Leokris [45]
2 years ago
5

Help..............................

Mathematics
1 answer:
nikdorinn [45]2 years ago
4 0

Answer:

2 hours

Step-by-step explanation:

Well you can make an equation for each

Lets “x” be hours

Jada, 28 +20x

Mai, 44 + 16x

So you can either make a table or a graph

I’m gonna make a table to make things easier

So lets do

jada: (0) 28  (1) 48  (2) 66  (3) 96

Mai: (0) 44  (1) 60  (2) 76  (3) 92

So we’re asked when is jada‘s service be cheaper

After 2 hours

jada <mai

whilst before

jada > mai

So 2 hours

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the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.


\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}


\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill

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2 years ago
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The answer to this question is C.
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3 years ago
Pls help have to turn in in like 5 minutes ​
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Step-by-step explanation:

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2 years ago
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Answer:

x = \frac{4-2y+2z}{4}

Step-by-step explanation:

4x-y+2z = 8x+y-4

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4 0
2 years ago
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Answer:

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Step-by-step explanation:

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You have a triangle with sides of 5 and 5, so that means the third side is:

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rounding:

5\sqrt{2} = 7

5 0
3 years ago
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