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2 years ago

If 17 ounces of cocoa are needed for 6 cookies, how many ounces of cocoa is needed for 8 cookies?

Mathematics

1 answer:

DENIUS [597]2 years ago

5 0

Answer:

22 and 2/3

Step-by-step explanation:

17/6= 2.8333333

2.8333333333 multiplied by 8= 22.66666666 or 22 and 2/3

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Brainliest if correct

n200080 [17]

Answer:

16 and 125

Step-by-step explanation:

-2*-2 = 4

4*-2 = -8

-8*-2 = 16

5*5 = 25

25*5 = 125

6 0

2 years ago

Triangle: solve for x

NARA [144]

Answer:

6 root two

Step-by-step explanation:

geometric mean for leg t.

24*3=x2

x2=72

x=root(72)

6 root two

if my answer helps please mark as brainliest.

6 0

3 years ago

Travis and his three friends ate out at Applebee's. Their bill totaled $72.50. If they left the server a 22%

Vlad [161]

The answer is a tip =15.95

8 0

2 years ago

Which best describes reduced fractions that will always terminate?

lesya692 [45]

Answer:Which fraction is a terminating decimal when written in decimal form?

Step-by-step explanation:1+1=2

4 0

2 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