Question

The function T(t) = 3t + 1, where 0 14 years ago

Answer 1

integral is 33/2
divide by 3 get 11/2

Answer 2

wow before i was so rudely interrupted.
\[\frac{1}{3}\int_0^3 3t+1 dt\]

Answer 3

what do you mean satellite?

Answer 4

you want average value yes?

Answer 5

average value of a function over an interval [a,b] is
\[\frac{1}{b-a}\int_a^b f(t)dt\]

Answer 6

in your case noon is 0, three o'clock is 3. average is
\[\frac{1}{3}\int_0^3 (3t+1)dt\]

Answer 7

the integral is easy to solve. it is
\[\frac{33}{2}\]

Answer 8

divide by 3 give
\[\frac{11}{2}\]so average is 5.2 must have been a cold day

Answer 9

sorry i mean 5.5

Answer 10

now the second part of the question asks when the temperature was in fact 5.5 so solve for t
[\3t+1=5.5\]
\[3t=4.5\]
\[t=1.5\]not coincidentally half way between 0 and 3

Answer 11

half way because you have a line

Answer 12

The animosity is so strong amongst these comments. Thanks.

Answer 13

didn't mean any sorry

Answer 14

the 'before i was rudely interrupted" part is because the site crashed right as i was typing. it was out for an hour or so

Answer 15

Oh. It was just a misunderstanding (: Thanks again