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2 years ago

$6.00 × 20 = $120 in 40 hours last week

Mathematics

1 answer:

BigorU [14]2 years ago

7 0

Answer:

If you are looking for the hourly wager, it is $3 per hour for 40 hours total.

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Maureen rolls a fair dice and flips a fair coin.

Taya2010 [7]

1/6 times 1/2 = 1/12

5 0

2 years ago

Need answer asap, correct answer will get brainliest

Cloud [144]

Answer:

x+5

Step-by-step explanation:

When you move to the right or left , that's on the x axis. And since it's to the right, the x- coordinate becomes larger. Hope this helps!

6 0

2 years ago

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There are 36 people waiting in line for hayride only six people can ride on each wagon if each wagon is full how many wagons are

ANEK [815]

Answer: 6 wagons will be needed

Step-by-step explanation:

36/6=6

3 0

3 years ago

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Drag steps in the given order to evaluate this expression. (−5)(2)−2(−3)+3 ( -5 )( 2 ) - 2 ( -3 ) + 3

Snowcat [4.5K]

(−5)(2)−2(−3)+3 ( -5 )( 2 ) - 2 ( -3 ) + 3

=-10+6-30+6+3

=15-40

=-25

6 0

3 years ago

PLEASE HELP MEEEE HURRRY!!! :)

sammy [17]

Answer:

Option D

Step-by-step explanation:

We are given the following equations -

$\begin{bmatrix}-5x-12y-43z=-136\\ -4x-14y-52z=-146\\ 21x+72y+267z=756\end{bmatrix}$

It would be best to solve this equation in matrix form. Write down the coefficients of each terms, and reduce to " row echelon form " -

$\begin{bmatrix}-5&-12&-43&-136\\ -4&-14&-52&-146\\ 21&72&267&756\end{bmatrix}$ First, I swapped the first and third rows.

$\begin{bmatrix}21&72&267&756\\ -4&-14&-52&-146\\ -5&-12&-43&-136\end{bmatrix}$ Leading coefficient of row 2 canceled.

$\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ -5&-12&-43&-136\end{bmatrix}$ The start value of row 3 was canceled.

$\begin{bmatrix}21&72&267&756\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\\ 0&\frac{36}{7}&\frac{144}{7}&44\end{bmatrix}$ Matrix rows 2 and 3 were swapped.

$\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&-\frac{2}{7}&-\frac{8}{7}&-2\end{bmatrix}$ Leading coefficient in row 3 was canceled.

$\begin{bmatrix}21&72&267&756\\ 0&\frac{36}{7}&\frac{144}{7}&44\\ 0&0&0&\frac{4}{9}\end{bmatrix}$

And at this point, I came to the conclusion that this system of equations had no solutions, considering it reduced to this -

$\begin{bmatrix}1&0&-1&0\\ 0&1&4&0\\ 0&0&0&1\end{bmatrix}$

The positioning of the zeros indicated that there was no solution!

<u><em>Hope that helps!</em></u>

6 0

3 years ago