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brilliants [131]

3 years ago

13

The Waters family budget a maximum amount of $150 per week for groceries. Mr. Waters already spent $40. Write and solve an inequ

ality to determine how much more the Waters family can spend on groceries.

1 answer:

ivann1987 [24]3 years ago

8 0

150-40 less than or equal to $110 -

They can spend less than or equal to $110

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A plumber has a 36-in pipe he must cut it into two pieces so that one piece is 14 inches longer than the other how long is each

ioda

The longest piece is 25 inches and the shortest piece is 11 inches.

(36/2) - (14/2) = shortest piece
18 - 7 = 11 (shortest piece)

36 - 11 = 25 (longest piece)

25 - 11 = 14 inches difference between the pieces

3 0

3 years ago

Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

Serjik [45]

Answer by JKismyhusbandbae:

|x| - 2 > 0

|x| -2 + 2 > 0 + 2

|x| > 2

Answer: x < - 2 or x > 2

4 0

3 years ago

Read 2 more answers

So what is the answer?

Dimas [21]

Answer for what?????

4 0

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.

5 0

3 years ago

What is the perimeter =26

UNO [17]

It depends upon the shape. But perimeter in general is the sum of the outside measure. Here, for whatever the shape is, the perimeter is 26

6 0

3 years ago