All categories

3 years ago

PLEASE HELP NO LINKS

Mathematics

1 answer:

Artyom0805 [142]3 years ago

7 0

Answer:

y = -2/3x + 3

Step-by-step explanation:

( -3, 5)

Parallel slope = -2/3

Slope-intercept:

y - y1 = m(x - x1)

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

y - 5 = -2/3x - 2

y = -2/3x + 3

You might be interested in

Points (−7, 5) and (7, 5) on the coordinate grid below show the positions of two players of a football team:

Greeley [361]

Visually, we can see that Player 1 and Player 2 are at the same vertical point, and we can confirm that numerically - both of their y-coordinates are 5. We can also see that it looks like the points mirror each other horizontally; numerically, their x-coordinates are 7 and -7 - the same basic number given different signs.

To find the axis of reflection, we want to find the line that passes right through the middle of the two points which, in this case, is the y-axis. Option 3 is the appropriate response in that light.

6 0

3 years ago

Read 2 more answers

What is the difference?

Crazy boy [7]

Answer:

<h2>D. $\frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\$ or StartFraction x squared + 3 x minus 12 Over (x + 3) (x minus 5) (x + 7) EndFraction</h2>

Step-by-step explanation:

Given the expression $\frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }$ , the dfference is expressed as follows;

Step1: First we need to factorize the denominator of each function.

$\frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }\\= \frac{x}{x^{2}-5x+3x-15 } - \frac{4}{x^{2} + 7x-5x - 35 }\\= \frac{x}{x(x-5)+3(x-5) } - \frac{4}{x( x+ 7)x-5(x +7) }\\= \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\\\$

Step 2: We will find the LCM of the resulting expression

$= \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\= \frac{x(x+7)-4(x+3)}{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+7x-4x-12 }{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\$

The final expression gives the difference

6 0

3 years ago

On a number line, the coordinates of A, B, C, and Dare -6, -2, 3, and 7, respectively. Find

den301095 [7]

Answer:

9: not congruent AB=-8 and CD=10

10: not congruent AC=-3 and BD=5

11: congruent BC=1 and AD=1

5 0

3 years ago

Heyy i need help on this maths q

Vikki [24]

Answer:

:) :D

Step-by-step explanation:

6 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

(a) x=2, y=-1

(b) x=2, y=2

(c) $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) x=-2, y=-7

Step-by-step explanation:

Cramer's Rule

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago