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Flauer [41]
3 years ago
12

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.
marin [14]

Answer:

the median is 43

Step-by-step explanation:

5 0
2 years ago
What is 2/5 times 15/16
Flauer [41]
That is 30/80 that reduces down to 3/8
How u get it is 2 times 15 over 5 times 16 then divide 30 and 80 by 10 to get 3/8
6 0
3 years ago
Read 2 more answers
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

Area = 744

744 = 62 * x

Divide both sides by 62

744/62 = 12

12 = x

That means the width would equal 12

12 * 62 = 744

5 0
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

-(infinity sign) is less than or equal to X is less than or equal to (infinity sign)
8 0
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
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