What is an equation of the line that passes through the point (−2,−3) and is parallel to the line 4x-y=14

Mathematics

1 answer:

liraira [26]3 years ago

5 0

Answer:

4x - y = -5

Step-by-step explanation:

The equation of a new line which is parallel to old line 4x - y = 14 looks exactly the same as that of the old line EXCEPT that we replace the constant, 14, with C and find C for the line that passes through (-2, -3).

4x - y = 14 becomes 4x - y = C

Now substitutte -2 for x, -3 for y and calculate C:

4(-2) - (-3) = C, or

-8 + 3 = C = - 5

Thus, the new equation is 4x - y = -5

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A pizzeria sells two types of pizza for lunch: pepperoni and cheese. The pepperoni pizza costs $8.50 and the cheese pizza costs

Usimov [2.4K]

Answer:

Let the no. Of cheese pizzas be x

Let the no. Of pepperoni pizzas be y

x + y = 45 (No. Of pizzas)

x = 45 - y

6x + 8.5y = 365 (cost of pizzas)

6(45-y) + 8.5y = 365

270 - 6y + 8.5y = 365

2.5y = 365 - 270

2.5y = 95

y = 95/2.5

y = 38

x = 7

Therefore 7 cheese pizzas were sold and 38 pepperoni pizzas were sold

Hope this helps!

7 0

3 years ago

A= [1, 3; 2, 1], B=[3, 6; -1, 1]. Find AB & BA if possible

steposvetlana [31]

Answer:

$AB\Rightarrow \quad \begin{bmatrix}0 & 9\\ 5 & 13\end{bmatrix}\\BA\Rightarrow \quad \begin{bmatrix}15 & 15\\ 1 & -2\end{bmatrix}$

Step-by-step explanation:

For two matrix P and Q, the product, say PQ is defined when:

The number of columns of P = The number of rows of Q

Since A is a 2×2 matrix and B is also a 2×2 matrix

Thus both AB and BA are possible.

So AB is:

$AB\Rightarrow\begin{bmatrix}1 & 3\\ 2 & 1\end{bmatrix}\begin{bmatrix}3 & 6\\ -1 & 1\end{bmatrix}\\AB\Rightarrow\quad \begin{bmatrix}3\times 1+3\times (-1) & 6\times 1+3\times 1\\3\times 2+1\times (-1) & 6\times 2+1\times 1\end{bmatrix}\\AB\Rightarrow \quad \begin{bmatrix}0 & 9\\ 5 & 13\end{bmatrix}$

BA is:

$BA\Rightarrow\begin{bmatrix}3 & 6\\ -1 & 1\end{bmatrix}\begin{bmatrix}1 & 3\\ 2 & 1\end{bmatrix}\\BA\Rightarrow\quad \begin{bmatrix}3\times 1+6\times 2 & 3\times 3+6\times 1\\(-1)\times 1+1\times 2 & (-1)\times 3+1\times 1\end{bmatrix}\\BA\Rightarrow \quad \begin{bmatrix}15 & 15\\ 1 & -2\end{bmatrix}$