A race car driver is doing a time trial of a new car on a straight track. The car's distance from the timekeeper is represented by y=293x+50, where x is time in seconds and y is distance in feet from the timekeeper's position. How many feet from the timekeeper is the race car at the beginning? It is ft from the timekeeper.

The distance at which the timekeeper is the race car at the start is 50 feet.

Step-by-step explanation:

You know that the car's distance from the timekeeper is represented by

y=293*x +50

where x is time in seconds and y is distance in feet from the timekeeper's position.

You want to determine how far the timekeeper is from the race car at the start. That is, the distance the timekeeper is from the car when the time is equal to zero.  This indicates that x = 0. Replacing x by that value in the expression of the distance of the car from the timekeeper as a function of time and solving:

y=293*0 +50

you get:

y=50

The distance at which the timekeeper is the race car at the start is 50 feet.

f(x) is the value of the function. b is the value of the function when x equals zero or the y-coordinate of the point where the line crosses the y-axis in the coordinate plane.

step-by-step explanation:

hope this !

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet