Answer:
The change in the car's distance is 8 feet
Step-by-step explanation:
* Lets explain how to solve the problem
- A car is driving away from a crosswalk
- The distance d (in feet) of the car from the crosswalk t seconds
since the car started moving is given by the formula d = t² + 3.5
- The time increasing from 1 second to 3 seconds
- We need to now the change of the car's distance from the crosswalk
∵ The equation of the distance is d = t² + 3.5
∵ The time is 1 second
∴ d = (1)² + 3.5
∴ d = 1 + 3.5 = 4.5 feet
∵ The time is 3 seconds
∴ d = (3)² + 3.5
∴ d = 9 + 3.5 = 12.5 feet
∵ The change of the distance = d of 3 sec - d of 1 sec
∵ d of 3 sec = 12.5 feet
∵ d of 1 sec = 4.5 feet
∴ The change of the distance = 12.5 - 4.5 = 8 feet
∴ The change in the car's distance is 8 feet
Area = l•w
length= 2x+1 width= x
210 = (2x+1)(x)
distribute x
210 = 2x^2 + x
move 210 to other side
2x^2+x-210 = 0
factor the polynomial
(2^x2+21x)-(20x-210)=0
x(2x+21)-10(2x+21)=0
(x-10)(2x+21)=0
x=10, -21/2
measurements can’t be negative so x=10
the width is 10
the length is 2(10)+1=21
Experimental probability is the ratio of number of times the event is occurring to the total number of the trials of the experiment. This implies that it is a ratio of the number of times an event is occurring to the total number of times that the activity has been repeated. Thus we shall use the formula:
Experimental probability=(# of occurrence of the event)/(total # of trials made)
=6/20
=3/10
Answer: 3/10
Answer:
for question 1 the answer is
Step-by-step explanation:
1)it would be 78.08-64 = 14 soo the precent is 14
Answer:
true
Step-by-step explanation:
