So, how far is the car from where it was at t = 0 is 40 m
<h3>Velocity of the car</h3>
Since the location x of the car in meters is given by the function x = 30t - 5t² where t is in seconds, we need to find the time at which its velocity is 10 m/s in the negative direction by differentiating x with respect to t to find its velocity, v.
So, v = dx/dt
= d(30t - 5t²)/dt
= d30t/dt - d5t²/dt
= 30 - 10t
When v is 10 m/s in the negative direction, v = -10 m/s.
So, v = 30 - 10t
-10 = 30 - 10t
-10 - 30 = -10t
-40 = -10t
t = -40/-10
t = 4 s
<h3>The distance at 4 s when its velocity is -10 m/s</h3>
Since at t = 4 s, its velocity is -10 m/s and x = 30t - 5t² is the car's location. The car's distance from t = 0 after its velocity is -10 m/s is
x(4) - x(0) = 30(4) - 5(4)² - [30(0) - 5(0)²]
= 120 - 5(16) - [0 - 0]
= 120 - 80 - 0
= 40 m
So, how far is the car from where it was at t = 0 is 40 m
Learn more about distance of car here:
brainly.com/question/17097458
Answer:
1/11
Step-by-step explanation:
If there are 11 counters and there's only 1 white one, that makes it to where there's only a 1 in 11 chance of you choosing that one white counter.
-4x + 8y = 2
-4x + 4x + 8y = 2 + 4x
8y = 4x + 2
8 8
y = ¹/₂x + ¹/₄
M>5=m>4
m>5 and m>7 are on line then m>5+m>7=180
by substituting
m>4+37=180
m>4=143
m>2=m>7=37
m>4=m>8
m<3+m>5=180
Given:
Box A has a surface area of 98 in².
Box B is 4 times the size of box A.
To find:
The surface area of box B.
Solution:
Let "a" be the side of box A , then "4a" be the side of box B because box B is 4 times the size of box A.
We know that the areas of similar figures are proportional to the square of their corresponding sides.




Using cross multiplication, we get


Therefore, the surface area of box B is 1568 in².