the formula y+20.25=4.58x + 4.50 models the stopping distance y in feet of a car traveling at pmh. estimate the distance traveled at 15 mph

[eddy2017]

hi, dear tutors!
i need a hint to start me off on this:
the formula y+20.25=4.58x + 4.50, models the stopping distance y in feet, of a car traveling at pmh. Estimate the distance traveled at 15 mph.
the first time i deal with this type of exercise.
Thanks in advance for any help,
eddy

 

[eddy2017]

rectifying an omission!!! traveling at x pmh!
i need a hint to start me off on this:
the formula y+20.25=4.58x + 4.50, models the stopping distance y in feet, of a car traveling at x pmh. Estimate the distance traveled at 15 mph.
the first time i deal with this type of exercise.
Thanks in advance for any help,

 

[Dr.Peterson]

rectifying an omission!!! traveling at x pmh!
i need a hint to start me off on this:
the formula y+20.25=4.58x + 4.50, models the stopping distance y in feet, of a car traveling at x mph. Estimate the distance traveled at 15 mph.
the first time i deal with this type of exercise.
Thanks in advance for any help,

You are to estimate the distance traveled (that's y, right?) if the car is going 15 mph (that's x, right?).

So plug in the given value for x, and solve for y.

(The wording could be better, and the formula is totally fake, but we take what we get.)

 

[eddy2017]

y= distance traveled=?
x= (distance traveled when the car is going at 15 mph)=15mph
let's solve for y
y+20.25=4.58x + 4.50
y + 20.25=4.58(15) + 4.50
y+20.25=73.2
y+20.25-20.25=73.2-20.25
y=52.95
distance traveled at 15 mph is ≈ 52.95 ft

 

[eddy2017]

thanksss!