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3 years ago

Let f(x) = (3x4 - 12)3 and h(x) = x3. Given that f(x) = (hºg)(x), find g(x).

Mathematics

1 answer:

blagie [28]3 years ago

6 0

Answer:

Step-by-step explanation:

f(x) = (3x4 - 12)3 and h(x) = x3.

g(x) = 3x4 - 12

since f(x) = (hºg)(x) = h(g(x))

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Camp counselors kept track of the number of granola bars they served each day.

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Answer:

the median is 43

Step-by-step explanation:

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2 years ago

What is 2/5 times 15/16

Flauer [41]

That is 30/80 that reduces down to 3/8
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3 years ago

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How do you find the area of the missing side

Margaret [11]

Lets use a rectangle as the example because I dont know what kind of shape you're looking for.

Area of a rectangle = length * width

Length (l) = 62

Width (w) = x

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3 years ago

Find the domain of the equation

goblinko [34]

The domain is all real numbers, since x can be any number and create a y value

This can be written in a few ways, including:

X:X

X: R

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3 years ago

hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and

Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0

1 year ago