The braking distance is the distance the car travels before coming to a stop after the brakes are applied
a. The braking distances are as follows;
- The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
- The braking distance at 55 mph, is approximately <u>298.35 ft.</u>
- The braking distance at 85 mph, is approximately <u>708.92 ft.</u>
b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s
Reason:
The given function for the braking distance is D = 2.6 + v²/22
a. The braking distance if the car is going 25 mph is therefore;
25 mph = 36.66339 ft./s
At 25 mph, the braking distance is approximately <u>63.7 ft.</u>
At 55 mph, the braking distance is given as follows;
55 mph = 80.65945 ft.s
At 55 mph, the braking distance is approximately <u>298.35 ft.</u>
At 85 mph, the braking distance is given as follows;
85 mph = 124.6555 ft.s
At 85 mph, the braking distance is approximately <u>708.92 ft.</u>
b. The speed of the car when the braking distance is 450 feet is given as follows;
v² = (450 - 2.6) × 22 = 9842.8
v = √(9842.2) ≈ 98.211 ft./s
The car was moving at v ≈ <u>98.211 ft./s</u>
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The range is {-37,-25,-13,-1}. So you need to figure out what four numbers from this list of numbers (1,2,3,4,5,6,7,8), when applied to this
function, ( f(x)=-6x+11 ), equals these numbers that are in the range {-37,-25,-13,-1}.
So you apply each of these numbers (1,2,3,4,5,6,7,8) into the function (f(x)=-6x+11)
one by one.
f(1)=-6(1)+11=5
f(2)=-6(2)+11= -1
f(3)=-6(3)+11= -7
f(4)=-6(4)+11= -13
f(5)=-6(5)+11= -19
f(6)=-6(6)+11= -25
f(7)=-6(7)+11= -31
f(8)=-6(8)+11= -37
As you can see, f(2),f(4),f(6),and f(8) equal the numbers that are in the range {-37,-25,-13,-1}.
She bought 5.5 pounds of cherries
explanation-
88 divided by 16 would be 5.5
Answer:
E. y - 4 = 6(x+2) is the answer
slope:
m = -8-4/-4+2 = 6
Formula:
y - ys1 = m(x - xs1)
hope this helps!
Step-by-step explanation:
Answer:
6w-60x+6
Step-by-step explanation:
6(w-10x+1)
6w+6(-10)+6*1
6w-60x+6