<h3>
Answer: False</h3>
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Explanation:
I'm assuming you meant to type out
(y-2)^2 = y^2-6y+4
This equation is not true for all real numbers because the left hand side expands out like so
(y-2)^2
(y-2)(y-2)
x(y-2) .... let x = y-2
xy-2x
y(x)-2(x)
y(y-2)-2(y-2) ... replace x with y-2
y^2-2y-2y+4
y^2-4y+4
So if the claim was (y-2)^2 = y^2-4y+4, then the claim would be true. However, the right hand side we're given doesn't match up with y^2-4y+4
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Another approach is to pick some y value such as y = 2 to find that
(y-2)^2 = y^2-6y+4
(2-2)^2 = 2^2 - 6(2) + 4 .... plug in y = 2
0^2 = 2^2 - 6(2) + 4
0 = 4 - 6(2) + 4
0 = 4 - 12 + 4
0 = -4
We get a false statement. This is one counterexample showing the given equation is not true for all values of y.
Answer:
3550
Convert to decimal.
0.7
Multiply 0.7
by 100
to convert to a percentage.
0.7⋅100
Simplify 0.7⋅100
.
70%
Step-by-step explanation:
Composing functions means that the input of the outer functions is the output of the inner function.
In fact, you can rewrite the circle notation as
So, we can substitute g(x) with its expression:
And since f(x)=x+5, we simply have to add 5 to its input:
Similarly, we have, substituting f with its expression,
And since g(x)=4x+2, we have to multiply the input by 4 and add 2:
![\sqrt[12]{2}](https://tex.z-dn.net/?f=%20%5Csqrt%5B12%5D%7B2%7D%20)