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alexandr1967 [171]
3 years ago
9

How do you write 4x+y=15 in function form?

Mathematics
1 answer:
sleet_krkn [62]3 years ago
3 0

Answer:

if you mean y-mx+b form

Step-by-step explanation:

y =− 4 x+15

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U = kx+y solve for x
zhuklara [117]

Step-by-step explanation:

Kx = u - y

Divide both sides by k

X = <u>u</u><u> </u><u>-</u><u> </u><u>y</u>

K

I think that's the answer

Hope it helps:)

6 0
3 years ago
Using the table below, what is Xavier's take-home pay?
damaskus [11]

Answer:

b

Step-by-step explanation:

6400+2448+1600+75=10523

32,000-10523=21477

7 0
3 years ago
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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
SOMEONE PLZZ HELP, I HAVE A TEST AND DONT KNOW HOW TO DO THIS!!
Harman [31]
I believe it is C.
Good luck!
5 0
3 years ago
How do I do number 4?
Galina-37 [17]
What are you trying to find?
4 0
3 years ago
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