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2 years ago

Somebody plz help me with this

Mathematics

1 answer:

Leviafan [203]2 years ago

6 0

Answer M=-2
Because when it’s negative it’s going from left to right and when it is to that means 2 up 1 to the right

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Helppppp !!!!!! 8^6 ÷ 8^7 A. 8^0 B. 8^1 C. 8^5 D. 8^-1 E. 8^10

MArishka [77]

$8^6 ÷ 8^7$ = $8^{6-7}$ = $8^{-1}$

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1 year ago

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What is 0.1265625 rounded to the nearest tenth

katrin2010 [14]

Answer:

0.1 is 0.1265525 rounded to the nearest tenth

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2 years ago

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Solve the systems algebraically please help :(

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2 years ago

Eliott was trying to factor 6x^2-18 . He found that the greatest common factor of these terms was 6 and made an area model:

Semenov [28]

x²-3

Step-by-step explanation:

The expression given is;

6x²-18

Finding the greatest common factor of the terms and simplifying the expression will be;

Divide the terms by 6 to get;

x²-3

Graphing the expression to find the area model will be as attached;

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Factorizing expressions:brainly.com/question/12290978

Keywords : factor, greatest common factor, terms, area model

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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!

Hope that helps!

6 0

2 years ago