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Zolol [24]
3 years ago
9

Which equation does not represent a function? A) y = 2x + 3 B) x = 4y + 7 C) 6x - 5y = 8 D) y2 = -5x - 8

Mathematics
2 answers:
yulyashka [42]3 years ago
6 0

Answer:

D) y2 = -5x - 8

Step-by-step explanation:

A function is a rule between two sets A and B so that each element of set A (Domain) corresponds to a single element of set B (Range). Mathematically they can be expressed as:

f: A\rightarrow B\\\\a\rightarrow b =f(a)

So:

y^2=-5x-8\\\\y=\pm \sqrt{-5x-8}

Isn't a function because the elements of the domain don't corresponds to a single element of the range.

We can see this in an easier way using the vertical line test, which is a visual way to determine if a curve is a graph of a function or not. If a vertical line intersects a curve in an xy plane more than once, then for a value of x the curve has more than one value of y, and then the curve does not represent a function.    

I attached you the graphs, so you can corroborate the results visually.

marin [14]3 years ago
4 0
D. It is not a function because there are multiple outputs for one input. 
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3 years ago
4 red balls 4 white balls what is the probability of drawing at least 1 red ball if two balls are drawn
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7 0
4 years ago
How to solve <img src="https://tex.z-dn.net/?f=log_%7By%7D%285y%29%20%3D%202" id="TexFormula1" title="log_{y}(5y) = 2" alt="log_
Natalija [7]

Answer:

  y = 5

Step-by-step explanation:

Expand the logarithm:

\log_y{(5)}+\log_y{(y)}=2\\\\\dfrac{\log{(5)}}{\log{(y)}}+1=2 \quad\text{change of base formula}\\\\\dfrac{\log{(5)}}{\log{(y)}}=1 \quad\text{subtract 1}\\\\\log{(5)}=\log{(y)} \quad\text{multiply by log(y)}\\\\5=y \quad\text{take the anti-log}

_____

You can also take the antilog first:

  5y = y²

  y(y -5) = 0 . . . . . subtract 5y, factor

  y = 0 or 5 . . . . . y=0 is not a viable solution, so y=5.

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4 years ago
What is the common ratio between successive terms in the sequence?
kobusy [5.1K]
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3 years ago
Read 2 more answers
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oksian1 [2.3K]

8.2is the answer

Step-by-step explanation:

|CB|+8

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