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scoray [572]
3 years ago
5

-3x2 (xy+2x2) + 7x4 -3x3(5y + x)??

Mathematics
1 answer:
allochka39001 [22]3 years ago
4 0

Answer:

−2x4−18x3y

Step-by-step explanation:

Let's simplify step-by-step.

(−3x2)(xy+2x2)+7x4−3x3(5y+x)

Distribute:

=(−3x2)(xy)+(−3x2)(2x2)+7x4+−3x4+−15x3y

=−3x3y+−6x4+7x4+−3x4+−15x3y

Combine Like Terms:

=−3x3y+−6x4+7x4+−3x4+−15x3y

=(−6x4+7x4+−3x4)+(−3x3y+−15x3y)

=−2x4+−18x3y

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The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in
Anon25 [30]

We have been given a quadratic function f(x)=(x-5)^{2} +2 and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is [2,\infty).

If we restrict the domain of this function to either (-\infty,5] or [5,\infty), it will become one to one function.

Let us know find its inverse.

y=(x-5)^{2}+2

Upon interchanging x and y, we get:

x=(y-5)^{2}+2

Let us now solve this function for y.

(y-5)^{2}=x-2\\
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Hence, the inverse function would be f^{-1}(x)=5+\sqrt{x-2} if we restrict the domain of original function to [5,\infty) and the inverse function would be f^{-1}(x)=5-\sqrt{x-2} if we restrict the domain to (-\infty,5].

8 0
3 years ago
Anybody...Help!!!!!!!!!!!
Sloan [31]
Simple...

when they ask for the rate of change they ask for slope -->>

Using \frac{ y_{1} - y_{2} }{ x_{1} - x_{2} } to find the rate of change for (-3,0) and (-2,-5)

\frac{0--5}{-3--2}

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Thus, your answer.


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Step-by-step explanation:

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