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3 years ago

Your cell phone company charges $20 per month for 500 minutes plus an additional

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

8 0

Answer:

$21

Step-by-step explanation:

$20 for the 500 minutes

We still have to pay for 10 minutes.

10 cents per minute.

10centsx10 minutes.

100 cents, also known as 1 dollar.

$20 plus $1 means the answer is $21.

Hope this helps!

-AxekickNebulite

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-3/2(x-2) simplify please

mario62 [17]

Answer:-3x+6/2

Step-by-step explanation:

To simplify -3/2(x-2)

We get

-3x+6/2

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3 years ago

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Which number is larger, 9.1 or √81 ? Explain

Katen [24]

Answer: 9.1

Sqrt(81)=9

So 9.1 is larger!
Thanks!

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3 years ago

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If these two shapes are similar, what is the measure of the missing length n?

scoundrel [369]

Answer:

39in.

Step-by-step explanation:

due to scale, cant be 15, as that would be the same length as other side

cant be 91 otherwise it wouldnt be similar if it had the same length as blue triangle which has 39 not 15 as the top side

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3 years ago

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.

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3 years ago

Given F(x) = 3x 2 - 4, find F(-√3). <br><br> a)-13<br> b)5<br> c)23

svet-max [94.6K]

I want to say 5 but I'm not sure

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3 years ago

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