All categories

3 years ago

Inverse of f(x)=3/2x +6

Mathematics

1 answer:

kvasek [131]3 years ago

7 0

Answer:

f(x) = (x - 6)/(3/2)

Step-by-step explanation:

You might be interested in

Is x = 7 a solution to the equation 3x = 21? yes no WHICH ONE??

Veronika [31]

Answer:

yes.

Step-by-step explanation:

if you plug 7 into x it'll be 3x7 and that will equal 21

4 0

3 years ago

Read 2 more answers

Simplify (7x - 7)(x + 1)

Nadusha1986 [10]

We will use the FOIL method to solve this.

(7x-7)(x+1)

Multiply 7x by x. Then, multiply 7x by 1. Then, multiply -7 by x. Then, Multiply -7 by 1.

$7x^{2} + 7x - 7x -7$

Now, we need to simplify by combining like terms.

$7x^{2} -7$

If you look, -7x and 7x are gone. This is because they cancel each other out.

I hope this helped and maybe consider giving me brainliest :)

5 0

3 years ago

Given that f(x) = 2x + 1 and g(x) = −5x + 2, solve for f(g(x)) when x = 3.

Anon25 [30]

The first step in solving for f(g(x)) when x=3, is solve for g(x). The answer of g(x) when x is equal to 3 is -(5) multiplied by 3 add by 2. Therefore, g(x) is -13. Then substitute the value of g(x) to f(x). The answer of f(-13) is 2 multiplied by -13, then add 1. So, the final answer is, f(x)=-25.

3 0

3 years ago

Read 2 more answers

Jamal has 1240 song on his mp3 player, and 40% of the songs are rock songs. how many are not rock songs?

Colt1911 [192]

Total songs = 1240
Rock songs = 40%
Songs that are not rock = 100 - 40 = 60%

It would be: 1240 * 60% = 1240 * 0.60 = 744

In short, Your Answer would be 744

Hope this helps!

4 0

3 years ago

Read 2 more answers

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago