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Novay_Z [31]
2 years ago
8

Question 24 Multiple Choice Worth 1 points)

Mathematics
1 answer:
hammer [34]2 years ago
8 0

Given:

The system of equations is:

Line A: y=x-4

Line B: y=3x+4

To find:

The solution of given system of equations.

Solution:

We have,

y=x-4              ...(i)

y=3x+4           ...(ii)

Equating (i) and (ii), we get

x-4=3x+4

-4-4=3x-x

-8=2x

Divide both sides by 2.

-4=x

Substituting x=-4 in (i), we get

y=-4-4

y=-8

The solution of system of equations is (-4,-8).

Now verify the solution by substituting x=-4, y=-8 in the given equations.

-8=-4-4

-8=-8

This statement is true.

Similarly,

-8=3(-4)+4

-8=-12+4

-8=-8

This statement is also true.

Therefore, (-4,-8) is a solution of the given system of equations, because the point satisfies both equations. Hence, the correct option is C.

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Its like a clock, except directly away from the six is not twelve, it's 13. That means the clock would have to have added two numbers for there to be a whole number facing the six.

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What is the volume of a sphere that has a diameter of 12
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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
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