PQ has endpoints at PH(7, -11) and Q(-5,-8). Find the slope of PQ

Simplify 3xy + x 2 - 4xy + 2xupper2y

densk [106]

The answer is X^2-xy+2x^2y sorry if I am wrong

6 0

3 years ago

Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x 2) f(x) = (x – 2)(x 2) f(x) = (x 1)(x – 2)

makvit [3.9K]

The function that has a vertex on the y-axis is f(x) = (x - 2)(x + 2)

<h3>How to determine the function?</h3>

For a function to have its vertex on the y-axis, then the coordinate of the vertex must be:

(h,k) = (0,y)

A quadratic function is represented as:

f(x) = (x - h)^2 + k

So, we have:

f(x) = (x - 0)^2 + k

Evaluate

f(x) = x^2 + k

From the list of options, we have:

f(x) = (x - 2)(x + 2)

Expand

f(x) = x^2 - 4

Hence, the function that has a vertex on the y-axis is f(x) = (x - 2)(x + 2)

Read more about vertex at:

brainly.com/question/18797214

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8 0

2 years ago

Help please!!!!!!!!!

In-s [12.5K]

ANSWER

24

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$

Since the determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

it follows that

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \\ &= (0)(-8) - (-3)(8) \\ &= -(-24) \\ &= 24 \end{aligned}$