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3 years ago

PLEASE HELP ME WITH THESE EXPONENTS! r^3 = w x w = 7 ^ 0 c ^ 1 =

Mathematics

1 answer:

mojhsa [17]3 years ago

8 0

Step-by-step explanation:

i juuts took this test

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What is the sum 5/3 +(1/5) please and explain

solmaris [256]

We need to find a common denominator(bottom number of fraction)

5/3 = 25/15 (multiply both top and bottom by 5)
1/5 = 3/15 (multiply both top and bottom by 3)

Now since the denominator is the same we can add to get 25/15 + 3/15 = 28/15

7 0

4 years ago

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kelly will roll a number cube labeled 1 to 6. what is the probability kelly will roll a number greater than 3?

guapka [62]

Answer:

%50

Step-by-step explanation:

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3 years ago

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I need help with this asap!!!!

Yuri [45]

Answer:

26.6 but that is not one of yor answers...

I think you have a typo in "A"

I think that your have a typo in the answers

x/sin(40) = 32/sin(90)

x = 26.6

Step-by-step explanation:

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3 years ago

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Can Someone plz help me with this question? I will mark brainliest

Gala2k [10]

The answer to your question is

No

8 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!

Hope that helps!

6 0

4 years ago