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At the curb a ramp is 11 inches off the ground. The other end of the ramp rests on the street 55 inches straight out from the curb. Write a linear equation in slope-intercept form that relates the height y of the ramp to the distance x from the curb.


y = 5x + 11

y = -1/5x + 11

y = -1/5x + 55

y = 1/5x + 55

Verified Answer :

Ckaranja

slope of the ramp=y/x
=11/55
=1/5
General equation;
y=mx+c
y=1/5x+c
Replacing for x and y;
11=1/5*55+c
c=0
The equation then should be;
y=1/5x

SociometricStar

Answer:

\(y=-\frac{1}{5}x+11\)

Step-by-step explanation:

The slope intercept form of a line is given by \(y=mx+b\)

Here m is the slope and b is the y intercept.

Now, the slope can be defined as the ratio of rise and run.

\(slope=\frac{rise}{run}\)

It has been given that at the curb a ramp is 11 inches off the ground. It means the rise is -11.

Now, other end of the ramp rests on the street 55 inches straight out from the curb. Hence, run is 55.

Therefore, the slope is given by

\(m=\frac{-11}{55}\\\\m=-\frac{1}{5}\)

Therefore, the linear equation is

\(y=-\frac{1}{5}x+b\)

Also, from the second condition, we have a point (55,0)

\(0=-\frac{1}{5}\cdot55+b\\0=-11+b\\b=11\)

Therefore, the linear equation is

\(y=-\frac{1}{5}x+11\)

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