All categories

3 years ago

What is the equation of the line through (-5,-1) & (3,-3)?

Mathematics

1 answer:

vivado [14]3 years ago

6 0

Answer:

y = -1/4x - 2.25

Step-by-step explanation:

If you do this algebraically then use formula:

M= (y2 - y1) / (x2 - x1)

In other words, M= (-3 - (-1)) / (3 - (-5))

M= (-2) / (8)

M= -1/4 <-- This is your slope

To find y-intercept:

Substitute M in the slope-intercept equation (y=mx+b) with -1/4, y=-1/4x+b

Next, Substitute y and x with either point (-5, -1) or (3,-3)

-1 = -1/4(-5) + b <-- Solve for B

-1.25 -1 = 1.25 + b -1.25

-2.25 = b

Now just substitute b and m, and there's your answer:

y = -1/4x - 2.25

Hope this helps!

You might be interested in

The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

3 years ago

68 - 4 × (6 - 4)4<br><br> pls help o-o

Alika [10]

Answer:

Here you go!, hope it helped. 68-32x

7 0

3 years ago

Read 2 more answers

What is the answer a, b, c, or d

podryga [215]

Answer: Choice B

(-2, 5)

==================================================

Explanation:

The original system is

$\begin{cases}-4x+3y = 23\\ x-y = -7\end{cases}$

Multiply both sides of the second equation by 3. Doing so leads to this updated system of equations

$\begin{cases}-4x+3y = 23\\ 3x-3y = -21\end{cases}$

Now add straight down

The x terms add to -4x+3x = -1x = -x

The y terms add to 3y+(-3y) = 0y = 0

The terms on the right hand sides add to 23+(-21) = 2

We end up with the equation -x = 2 which solves to x = -2

Now use this to find y. You can pick any equation with x,y in it

----------------

-4x+3y = 23

-4(-2)+3y = 23

8+3y = 23

3y = 23-8

3y = 15

y = 15/3

y = 5

x-y = -7

-2-y = -7

-y = -7+2

y = -5

y = 5

Either way, we get the same y value.

So that's why the solution is (x,y) = (-2, 5)

6 0

3 years ago

Which equation represents the line whose slope is 1/2 and whose y-intercept is 5

expeople1 [14]

Answer:

y=1/2x+5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Can anyone help don’t understand

makvit [3.9K]

Answer:

Twenty decreased by a number. The question says that "a number" =X.

twenty = 20

decreased = minus

by a number = X

Put this all in a line and you get...

an answer of:

20 - X.

3 0

2 years ago