Burgers-o-Rama is the best hamburger place in town. The owner, Ms. Hamm,

Y=x-3 and y=-3x+25 in substitution form

Vlad1618 [11]

(x-3)=-3x+25

Because we know that y=x-3 so we put that in place of y in the equation

8 0

3 years ago

Solve the system of elimination -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5

Dahasolnce [82]

Answer:

x = 1 , y = 1 and z = 0

Step-by-step explanation:

-2x+2y+3z=0 ------(1)

-2x-y+z=-3 --------(2)

2x+3y+3z=5 ---------(3)

To find the values of x,y and z

(3) - (2)⇒ 2y + 4z = 2

y + 2z = 1 --------(4)

(1) - (2)⇒ 3y +2z =3 ---(5)

(5) - (4)⇒ 2y = 2

y = 1

Substituting the value of y in (4) we get z =0

Substituting the value of y and z in (1) we get x = 1

Therefore x = 1 , y = 1 and z = 0

5 0

3 years ago

In the function y = a(x – h)2 + k

coldgirl [10]

Answer:

h is the horizontal shift

Step-by-step explanation:

y = a(x – h)^2 + k

This is the equation for a parabola in vertex form

The vertex is ( h,k) and it is shifted from (0,0)

h is the horizontal shift and k is the vertical shift

8 0

2 years ago

Read 2 more answers

Tom has some toy cars. 25 of them are red. He has 35 more green cars than red cars. The remaining 15 toy cars are yellow. Tom de

horrorfan [7]

Answer:

Step-by-step explanation:

red = 25

green = 25 + 35 = 60

yellow = 15

red = 25/5 = 5

green = 60/5 = 12

yellow = 15/5 = 3

Red + green = 5 + 12 = 17 boxes

yellow = 3 boxes

17 - 3 = 14 boxes

There are 14 more boxes of red and green cars than yellow cars.

8 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!

Hope that helps!

6 0

3 years ago