Question

:D

Answer 1

What do you need help with

Answer 2

well for question 12

35 is the centre of the curve... and 65 is the amplitude...

so the max temperature is 35 + 65

the period is

\[\frac{2\pi}{b}\]

in the question

\[b = \frac{\pi}{5}\]

so the period is

\[\frac{2\pi}{\frac{\pi}{5}}\]

jst simplify the fraction to find the period of the experiment

hope it helps

Answer 3

I was going to say what cambell was going to say

Answer 4

for question 13 just substitute t = 5 and then evaluate it.

question 14 doesn't make a lot of sense...

Answer 5

thats correct the period is 10

Answer 6

and max temp is 100

Answer 7

I didn't think you were asked for the min.... but -30 is correct...

Answer 8

oops... you were

Answer 9

well you have an equation with t as a variable... so replace t with the value of 5 and then calculate...

\[f(5) = 4\cos(\frac{\pi}{3} \times 5) + 15\]

Answer 10

you need to realise....

\[\frac{5\pi}{3}\]

is a 3rd quadrant angle and so it becomes

\[f(5) = 4 \times -\cos(\frac{2\pi}{3}) + 15\]

Answer 11

no you can just type it into the calculator as

\[-4 \times \cos(\frac{2\pi}{3}) + 15\]

Answer 12

17 seems to make sense...

Answer 13

for question 14 I think there is something missing...

as my simplified solution would be

\[(\cos(\theta) + \cos(\theta))^2 + (\cos(\theta)+\cos(\theta))^2 = 2(2\cos(\theta))^2 = 8\cos^2(\theta)\]

Answer 14

12 is D the period is 10 hours.... max 100 min -30

Answer 15

I have to go good luck