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Two points located on jk are j (-1,-9) and k (5,3). What is the slope of jk?

Answers

Answer 1
Answer:

Answer:

Slope = 2

Step-by-step explanation:

Slope = (rise)/(run)

Slope = (3+9)/(5+1)

Slope = (12)/(6)

Slope = 2
Answer 2
Answer:

In the given case, we can conclude that The slope of the line JK is 2.

To find the slope of the line that passes through the points J(-1,-9) and K(5,3), we can use the formula: slope = (y2 - y1) / (x2 - x1).

The slope of a line is a measure of how steep the line is. It describes the rate at which the dependent variable (usually denoted as 'y') changes with respect to a change in the independent variable (usually denoted as 'x').

Plugging in the coordinates, we get:

slope = (3 - (-9)) / (5 - (-1)) = 12 / 6 = 2.

Learn more about Slope here:

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Answers

Answer:

40.62

Step-by-step explanation:

that's the answer I got

What is the y value of the line when x = -1

Answers

I cannot answer that question because I need to equation

Which best describes how to find an equation of the line shown?A. The slope m is rise over run, so y over x =m. Solve for Y to get y=mx.

B. The slope m is run over rise, so x over y=m. Solve for x=ym.

C. The slope m is run over rise, so x over y=m. Solve for y to get y= x over m.

D. The slope m is rise over run, so x over y=m. Solve for x to get x=ym.

Answers

Answer:

A

Step-by-step explanation:

m (slope) =  rise    =    y    =  y₁ - y₁

                   run           x        x₁ - x₁

to get y = mx

therefore, the answer to the question is :

A. The slope m is rise over run, so y over x =m.

    Solve for y to get y=mx.

Find a solution to the following initial-value problem: dy dx = y(y − 2)e x , y (0) = 1.

Answers

This equation is separable, as

(\mathrm dy)/(\mathrm dx)=y(y-2)e^x\implies(\mathrm dy)/(y(y-2))=e^x\,\mathrm dx

Integrate both sides; on the left, expand the fraction as

\frac1{y(y-2)}=\frac12\left(\frac1{y-2}-\frac1y\right)

Then

\displaystyle\int(\mathrm dy)/(y(y-2))=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C

\implies\frac12\ln\left|\frac{y-2}y\right|=e^x+C

Since y(0)=1, we get

\frac12\ln\left|\frac{1-2}1\right|=e^0+C\implies C=-1

so that the particular solution is

\frac12\ln\left|\frac{y-2}y\right|=e^x-1\implies\boxed{y=\frac2{1-e^(2e^x-2)}}

This is very important assignment I need to make my family proud with a good grade

Answers

Answer:

They're 24 in total so i guess it would be 24 units^2

Step-by-step explanation:
24!!! i hope you make them proud :)

Someone help pls!!! Thanks

Answers

Answer:

The answers are A and C

Step-by-step explanation:

A is correct because -16 is on the left of the number line.
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