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Veseljchak [2.6K]
3 years ago
6

Which of the following relations is a function ?

Mathematics
1 answer:
Effectus [21]3 years ago
6 0

Answer:

Step-by-step explanation:

The relation that is a function is c

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PLZ HELP DUE AT 12
aleksandrvk [35]

Answer:

The first one would be m^30n^20p^30  

And the second one would be a^2 / d^4  

Step-by-step explanation:

For 1, you have to multiply the inner exponents by the outer one so you get (m^12n^8) times (m^18n^12p^30). Then you mutiply the two together, so you add the exponents, where you get m^30n^20p^30.

For 2, since you're dividing, you have to subtract the exponents so you get a^2d^-4, but since you shouldn't have the negative you turn it into a^2 / d^4

Hope this helps :)

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3 years ago
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We need an image to be able to answer this question

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3 years ago
Which of these limits evaluate to 0?
Vikentia [17]
<h3>Answer: C) I and II only</h3>

===============================================

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\\\\\\

----------

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\\\\\\

----------

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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3 years ago
PLEASE HELP THIS IS MY LAST QUSTION OF THE DAY!!! :)<br> thanks in advanced
notsponge [240]

So using standard form you convert the equation to 5x - 2y = -10 then use slope m of a line of the form Ax + By = C equals negative a over b. It would be a = 5 b = -2 m = -5/-2 m = 5/2. Have a good day the answer is 5/2

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3 years ago
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goldfiish [28.3K]

Answer:

Sorry, i dont know

Step-by-step explanation:

I dont know the answer to this question.....

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3 years ago
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