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neonofarm [45]
3 years ago
13

What is the equation of the line that passes through (4, -1) and (-2, 3)?

Mathematics
2 answers:
Zielflug [23.3K]3 years ago
6 0

Answer:

\displaystyle y=-\frac{2}{3}x+\frac{5}{3}

Step-by-step explanation:

We want to find the equation of the line that passes through the two points:

(4, -1) and (-2, 3).

First, we will find the slope of the two points. So:

\displaystyle m=\frac{3-(-1)}{-2-4}=\frac{4}{-6}=-\frac{2}{3}

Now, we will use the point-slope form given by:

y-y_1=m(x-x_1)

Use either point. I'm going to use (4, -1). So, substitute:

\displaystyle y-(-1)=-\frac{2}{3}(x-(4))

Simplify and distribute:

\displaystyle y+1=-\frac{2}{3}x+\frac{8}{3}

Subtract 1 from both sides. Therefore:

\displaystyle y=-\frac{2}{3}x+\frac{5}{3}

rewona [7]3 years ago
4 0

Answer:

3--1/-2-4=4/-6=2/-3

y=-2/3x+b

-2*-2/3=4/3

3-4/3=5/3

y=-2/3x+5/3

Step-by-step explanation:

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Find the length of X ( IN THE PICTURE)​
vladimir2022 [97]

Answer:

x=3

Step-by-step explanation:

The triangles are similar so we can use ratios to solve

x           4

----- = -----------

4.5        6

Using cross products

6x = 4.5 * 4

6x =18

Divide by 6

6x/6 = 18/6

x = 3

3 0
3 years ago
Solve 3x2 + 6x = 8.
-Dominant- [34]

Answer:

The quantity of negative 1 plus or minus square root 33 all over 3

Step-by-step explanation:

3x2+6x=8

3x^2+6x-8=0

Quadratic formula

x=-b +or-√b^2-4ac/2a

a=3

b=6

c=-8

x=-b +or-√b^2-4ac/2a

=-6 +or- √(6)^2 - 4*3*-8/2*3

=-6 +or- √36-(-96)/6

=-6 +or-√36+96/132

=-6 +or- √132/6

=-6 +or- 2√33/6

=-6/6 +or- 2√33/6

= -1 +or- √33/3

The quantity of negative 1 plus or minus square root 33 all over 3

8 0
4 years ago
Read 2 more answers
Please help me! I will give Brainly
dolphi86 [110]

Answer:

8

Step-by-step explanation:

6^7 : 6^6 = 6^(7-6) = 6^1 = 6

3^3 : 3^4 = 3^(3-4) = 3^-1 = 1/3

6* 1/3 = 2

2^3 = 8

4 0
3 years ago
Help! please i cant figure it out.
ki77a [65]

let's recall that the graph of a function passes the "vertical line test", however, that's not guarantee that its inverse will also be a function.

A function that has an inverse expression that is also a function, must be a one-to-one function, and thus it must not only pass the vertical line test, but also the horizontal line test.

Check the picture below, the left-side shows the function looping through up and down, it passes the vertical line test, in green, but it doesn't pass the horizontal line test.

now, check the picture on the right-side, if we just restrict its domain to be squeezed to only between [0 , π], it passes the horizontal line test, and thus with that constraint in place, it's a one-to-one function and thus its inverse is also a function, with that constraint in place, or namely with that constraint, cos(x) and cos⁻¹(x) are both functions.

7 0
2 years ago
Graph this Function 1: f(x) = −3x2 + 2
jasenka [17]

I guess that's f(x) = -3x^2+2

Well :

-3x^2+2=0

-3x^2=-2 / : (-3)

x^2= -2/-3

x^2= 2/3 / sqrt

x= +/ - sqrt(2)/sqrt(3)

We got the zero points.

x ~ +/- 0.81

f(0)= -3(0)^2 + 2

f(0)= 2

Mark three points:

(0.81;0)

(-0.81;0)

(0;2)

Connect with parabolic line and ya got a your graph

3 0
4 years ago
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