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3 years ago

What is the vaule or r + t when r= -12, and t =-8

Mathematics

2 answers:

Maurinko [17]3 years ago

4 0

The answer is -20 for real

Tems11 [23]3 years ago

3 0

The answer to your question is -20 i hope this helps!

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Graph the line.<br> y + 4x=8

const2013 [10]

Y-Intercept =(−8)
Go to the graph and plot the point the point (0,−8) on the y-axis.
Then rise 4 and run 1. Plot the point (1,−4)
Then again "rise 4" and then "run" 1. Plot the point (2,0)

6 0

3 years ago

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Plz plz plz help I don’t get this

Bumek [7]

It looks like it counts up by 2 when adding. so add 4, then add 6, then add 8, then add 10, etc. to find the output you simply add 2 more than the last time you added, to find the input, you subtract 2 more than last time. hope this makes sense haha.

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3 years ago

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How many tires are there if you have 10 bicycles?

mash [69]

20 bicycles. two tires per one bike. 2*10=20

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2 years ago

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Hii pls help i’ll give you brainliest if you give a correct answer.

Maksim231197 [3]

Answer:

19.375

Step-by-step explanation:

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2 years ago

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Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

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Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

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Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$

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Part III

$\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\$

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2 years ago