Write an equation in slope-intercept form (y=mx+b) for the line that has a slope of 12 and passes through the point (3, 20).

y=12x-16

Step-by-step explanation:

You want to find the equation for a line that passes through the point (3,20) and has a slope of 12.

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

To start, you know what m is; it's just the slope, which you said was 12. So you can right away fill in the equation for a line somewhat to read:

y=12x+b.

Now, what about b, the y-intercept?

To find b, think about what your (x,y) point means:

(3,20). When x of the line is 3, y of the line must be 20.

Because you said the line passes through this point, right?

Now, look at our line's equation so far: . b is what we want, the 12 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the the point (3,20).

So, why not plug in for x the number 3 and for y the number 20? This will allow us to solve for b for the particular line that passes through the point you gave!.

(3,20). y=mx+b or 20=12 × 3+b, or solving for b: b=20-(12)(3). b=-16.

The equation of the line that passes through the point (3,20) with a slope of 12

is

y=12x-16

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.