Each side after the cut-out is (s-2c) where s is the original side length and c is the size of the cut so:
V=hlw, h=x and l and w are both (x-8)
V=x(x-8)(x-8)
V=x(x^2-16x+64)
V=x^3-16x^2+64x
Answer:
is this actually a question
Step-by-step explanation:
Given: <span>9=(1/27)^(a+3)
[Note: Please review rules of PEMDAS.
The question posted as is means
</span><span>9=[(1/27)^a]+3 which is probably not what you mean. It has been interpreted with (a+3) as the exponent, please check.]
</span><span>
Need to find value of a that satisfies above.
We will first isolate a in order to solve for its value.
</span>9=(1/27)^(a+3)
Switch unknown to the left
(1/27)^(a+3) = 9
Since both 27 and 9 are powers of 3, we can take log to base 3.
take log (base 3) on both sides, applying rule of logarithm of exponents.
(a+3)log_3(1/27)=log_3(9)
(a+3)log_3(3^(-3))=log_3(3^2)
Apply definition of logarithm [log_a(a^k)=k]
(a+3)(-3)=2
(a+3)=-2/3
a=-3 2/3=-11/3
Check: (1/27)^(-11/3+3)=(1/27)^(-2/3)=(3^(-3))^(-2/3)=3^(-3(-2/3))=3^2=9 ok
Answer: a=-11/3 or a=-3.667 (approx.)
Answer:
g(x) = f(3x) -15
Step-by-step explanation:
Let f(x) be the parent function.
If the function is to be compressed by a factor of 3 horizontally, we must obtain the value f(3) when x=1, for example. We can do that by using 3x as the argument of the function:
g(x) = f(3x)
If its graph is to be shifted downward by 15 units, every value produced by the function must have -15 added to it. That is, the shifted function is ...
g(x) = f(3x) - 15
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The attached graph shows an example function compressed horizontally and vertically shifted.