Write the slope intercept form of the equation through the given point with the given slope.

Write the equation of the line that passes through (3, 4) and (2, −1) in slope-intercept form. (2 points) a y = 3x − 7 b y = 3x

sashaice [31]

Answer: y = 5x − 11

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis represent

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

The line passes through (3,4) and (2, -1),

y2 = - 1

y1 = 4

x2 = 2

x1 = 3

Slope,m = (- 1 - 4)/(2 - 3) = - 5/- 1 = 5

To determine the y intercept, we would substitute x = 3, y = 4 and m= 5 into

y = mx + c. It becomes

4 = 5 × 3 + c

4 = 15 + c

c = 4 - 15 = - 11

The equation becomes

y = 5x - 11

8 0

3 years ago

Simplify expression X^2+x-2/x^3-x^2+2x-2

professor190 [17]

Answer:

$=\frac {x+2}{x^2+2}$ is the simplest form of given expression.

Step-by-step explanation:

The given question is $\frac{x^2+x-2}{x^3-x^2+2x-2}$

To solve the problem we have to group or split middle term and then factorise

$\frac{x^2+(2x-x)-2}{(x^3-x^2)+(2x-2)}$

Taking $x^2$ common from first two terms of denominator and 2 from next two terms

$= \frac{x^2+2x-x-2}{x^2(x-1)+2(x-1)}$

Now,taking x common from first two terms of numerator and -1 from next two terms and in denominator taking(x-1) common from both terms

$= \frac{ x(x+2)-1(x+2)}{(x-1)(x^2+2)}$

$=\frac{(x+2)(x-1)}{(x-1)(x^2+2)}$

Now cancel out x-1 from both numerator and denominator we get

$=\frac {x+2}{x^2+2}$ is the required simplest form.

7 0

3 years ago

Which statement is true regarding DC current?

DerKrebs [107]

I think it's this answer <span>C.A high voltage current can be transformed into a low voltage current.</span>

6 0

3 years ago

Read 2 more answers

Just a tad bit more points :)

swat32

Answer:

Thx have a great day :)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel

nydimaria [60]

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

$\theta=\frac{\pi}{3}$

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be= $r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j$

By using $cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

Force along the plane will be= $\mid F_x\mid=F\cdot r$

Force along the plane will be = $\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3$ N

By using $i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0$

Therefore, force along the plane= $\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)$

2.The vector perpendicular to the plane= $r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j$

The force perpendicular to the plane= $\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

The force perpendicular to the plane= $4$ N

Therefore, $F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $2\sqrt 3i-6j-2\sqrt3 i-2j=-8j$

Hence,sum of two component of forces=Total force.

6 0

3 years ago