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2 years ago

If y = (x-5)(x+7) then what does x equal?

Mathematics

1 answer:

DochEvi [55]2 years ago

7 0

Answer:

x = 5, x = -7

Step-by-step explanation:

If you mean find the roots, then x = 5 and x = -7

Because roots makes the equation equal to 0 and 5 - 5 = 0 and -7 + 7 = 0

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Tina paid $6.60 for some $0.15 stamps and some $0.20 stamps. She bought 37 stamps in all. How many of each kind did she buy?

sergiy2304 [10]

Answer:

Let n be the number of 25 cent stamps.Then, 28 - n must be the number of 29 cent stamps.Tim paid $7.60 the stamps.

0.25n + 0.29(28 - n) = 7.60

This means that n stamps at 25 cents a piece plus (28 - n) stamps at 29 cents a piece sum to $7.60.Solve for m.

0.25n + 8.12 - 0.29n = 7.60

-0.04n = -0.52

n = 13

So, Tim bought 13 25-cent stamps and 28 - 13 = 15 29-cent stamps

Step-by-step explanation:

hope this helps:)

4 0

2 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:

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