Acar’s stopping distance in feet is modeled by the equation d(v)= 2.15v^2/58.4f where v is the initial velocity of the car in miles per hour and f is a constant related to friction. if the initial velocity of the car is 47 mph and f = 0.34, what is the approximate stopping distance of the car?
a. 21 feet
b. 21 miles
c. 239 feet
d. 239 miles

C is the correct answer.The approximate stopping distance of the car =239 feet.

Step-by-step explanation:

Given: A car’s stopping distance in feet is modeled by the equation

where v is the initial velocity of the car in miles per hour and f is a constant related to friction.

If the initial velocity of the car v= 47 mph and f = 0.34 then

The distance of the car

≈239 feet           [ round to nearest ones]

Therefore ,the approximate stopping distance of the car =239 feet.

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.