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OpenStudy (howard-wolowitz):
@geerky42
geerky42 (geerky42):
Other one said it's A, but I would say D.
I would try for D.
OpenStudy (howard-wolowitz):
how do you know the answers so quickly?
geerky42 (geerky42):
Seen this question before.
geerky42 (geerky42):
Actually it's B.
\(\dfrac{10h^2}{h} = 10h\)
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geerky42 (geerky42):
Oops A*
OpenStudy (howard-wolowitz):
which one this one or the first one
geerky42 (geerky42):
It's A.
OpenStudy (howard-wolowitz):
16 or 14
geerky42 (geerky42):
Oh. 14.
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geerky42 (geerky42):
D.
Switch y to x;
\[x = \sqrt{8y+1}\\x^2 = 8y+1\\x^2-1 = 8y\\\boxed{y = \dfrac{x^2-1}{8}}\]
OpenStudy (howard-wolowitz):
ok so 11. D
14. A
16. D
geerky42 (geerky42):
Yep.
OpenStudy (howard-wolowitz):
AWESOME
geerky42 (geerky42):
You got them all correct?
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OpenStudy (howard-wolowitz):
are you supruised
geerky42 (geerky42):
Not really, Just wanna check, especially for 16.
OpenStudy (howard-wolowitz):
how did you learn how to do all of this?
geerky42 (geerky42):
Mostly by practice lol.
geerky42 (geerky42):
If you put in effort to learn and keeping try, you will be good.
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geerky42 (geerky42):
More of less why you get homework. it's practice.
geerky42 (geerky42):
Well, \((f\circ g)(x) = f(~g(x)~)\), right?
So you have \((f\circ g)(-2) = f(~g(-2)~)\)
From graph, we can see that \(g(-2)= 0, right?\)
Now we have \(f(~g(-2)~)=f(0)=\boxed{-4}\)
Does that make sense?