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ra1l [238]
3 years ago
14

Just by looking at y = −x 2 + 1, how do you know that it does not represent a line?

Mathematics
2 answers:
bixtya [17]3 years ago
7 0

Answer:

well its because its -x2 this cannot be graphed

Step-by-step explanation:

alexgriva [62]3 years ago
5 0

Answer:

Step-by-step explanation:

Because it doesn’t have a symmetry line

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Helpppp!!!!!!!!!!!!!
SIZIF [17.4K]

Answer:

a

Step-by-step explanation:

7 0
3 years ago
[7.07] Choose the correct product of (6x + 2)2.
Elodia [21]
Answer:
36x^2 + 24x + 4

Explanation:
The perfect square has the following general formula:
(a + b)^2 = a^2 + 2ab + b^2

Now, we will expand the given perfect square based on the above general formula:
(6x + 2)^2 = (6x)^2 + 2(6x*2) + (2)^2
                 = 36x^2 + 24x + 4

Hope this helps :)
4 0
4 years ago
Read 2 more answers
Find the range of F<img src="https://tex.z-dn.net/?f=f%28x%29%3D%20%5Cfrac%7B2%7D%7Bx-7%7D%20" id="TexFormula1" title="f(x)= \fr
steposvetlana [31]
Answer: y \in R; y \neq 0
Explanation: For this, it is often best to find the horizontal asymptote, and then take limits as x approaches the vertical asymptote and the end behaviours.

Well, we know there will be a horizontal asymptote at y = 0, because as x approaches infinite and negative infinite, the graph will shrink down closer and closer to 0, but never touch it. We call this a horizontal asymptote.
So we know that there is a restriction on the y-axis.

Now, since we know the end behaviours, let's find the asymptotic behaviours.

As x  approaches the asymptote of 7⁻, then y would be diverging out to negative infinite.
As x approaches the asymptote at 7⁺, then y would be diverging out to negative infinite.

So, our range would be: y \in R; y \neq 0
6 0
3 years ago
Whats 14 to the 2nd power
ExtremeBDS [4]

Answer: 130.666

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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LenaWriter [7]
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5 0
4 years ago
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