The distance s (in feet) covered by a car t sec after starting is given by the following function. s = −t3 + 11t2 + 20t (0 ≤ t ≤ 6) find a general expression for the car's acceleration at any time t (0 ≤ t ≤ 6). s ''(t) = correct: your answer is correct. at what time t does the car begin to decelerate? (round your answer to one decimal place.) t = sec

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!

you already gave the answer

step-by-step explanation: