Question

The function f(t)=5cos(pi/4^t)+11 represents the tide in the Dark Sea. It has a maximum of 16 feet when time (t) is 0 and a minimum of 6 feet. The sea repeats this cycle every 8 hours. After 6 hours, how high is the tide?

Answer 1

A. 11 feet
B. 6 feet
C. 8.5 feet
D. 10.5 feet

Answer 2

you can plug in 6 into t

Answer 3

\[f(6)=5\cos (\pi/4^6)+11 l\]

Answer 4

like that?

Answer 5

thats a bit strange to have time as the exponent

Answer 6

Yeah I'm not really how to solve it now. The word problems on my homework are always hard.

Answer 7

$$ \Large{ f(t)=5\cos(\frac{\pi}{4}\cdot t)+11
\\ ~\\
f(6) = 5\cos(\frac{\pi}{4}\cdot 6)+11
}$$

Answer 8

How do I go about solving this?

Answer 9

we can use a calculator or our knowledge of the unit circle

Answer 10

what is cos (6/4 pi ) ?

Answer 11

simplify to
cos (3/2 pi )

Answer 12

Then what?

Answer 13

multiply that by 5 , then add 11

Answer 14

we can use google's calculator
https://www.google.com/search?q=5cos%286%2F4+*pi%29+%2B11&ie=utf-8&oe=utf-8

Answer 15

I got 11 feet

Answer 16

correct ☺

Answer 17

Could you help with 2 other questions please?

Answer 18

ok

Answer 19

Simplify (sinƟ-cosƟ)^2+(sinƟ+cosƟ)^2
A. -sin^2Ɵ
B. -cos^2Ɵ
C. 0
D. 2

Answer 20

i would expand that

Answer 21

(sinƟ-cosƟ)^2+(sinƟ+cosƟ)^2
=(sinΘ-cosΘ)(sin Θ - cos Θ ) + (sinΘ+cosΘ)( sin Θ + cos Θ)

Answer 22

Don't they all cross out?

Answer 23

not necessarily, did you foil it out?

Answer 24

One sec

Answer 25

I'm confused on how to use foil on it

Answer 26

(sinƟ-cosƟ)^2+(sinƟ+cosƟ)^2
=(sinΘ-cosΘ)(sin Θ - cos Θ ) + (sinΘ+cosΘ)( sin Θ + cos Θ)
= sin^2 Θ - cosΘ sin Θ - cosΘsin Θ + cos^2 Θ + sin^2 Θ + cosΘ sin Θ + cosΘ sin Θ + cos^2Θ

Answer 27

I got \[\sin ^2\theta-\cos^2\theta+\sin^2\theta+\cos^2\theta\]

Answer 28

(sinƟ-cosƟ)^2+(sinƟ+cosƟ)^2
=(sinΘ-cosΘ)(sin Θ - cos Θ ) + (sinΘ+cosΘ)( sin Θ + cos Θ)
= sin^2 Θ - cosΘ sin Θ - cosΘsin Θ + cos^2 Θ + sin^2 Θ + cosΘ sin Θ + cosΘ sin Θ + cos^2Θ
=sin^2 Θ - 2cosΘ sin Θ+ cos^2 Θ + sin^2 Θ +2cosΘ sin Θ + cos^2Θ
=2sin^2 Θ + 2cos^2 Θ
= 2 ( sin^2 Θ + cos^2 Θ )
= 2*1

Answer 29

Oh I understand it now. It's not that hard after it's explained lol

Answer 30

☺ ☺

Answer 31

So it's 2. I have one more question that I didn't understand

Answer 32

Compare the following functions. Which function has the smallest minimum?

Answer 33

Thats one of them hang on

Answer 34

f(x)= -5sin(2x-pi)+2

h(x)
x y
-2 10
-1 7
0 5
1 3
2 5
3 7
4 10

Answer 35

The graph is g(x)

Answer 36

@perl still there?

Answer 37

I think it is f(x)

Answer 38

what is the minimum of f(x)

Answer 39

I got -3

Answer 40

It looks like -2 to me

Answer 41

So they all have the same minimum?

Answer 42

wait, there are three functions?

Answer 43

Yes, the graph, the f(x) function and the table

Answer 44

ok the minimum of the function in the table is 3.
the minimum of the graph is -2
the minimum of the sine function is -5*1 + 2 = -3

Answer 45

So it is f(x)

Answer 46

correct :)

Answer 47

Yay! thank you so much! :)

Answer 48

I fanned :)

Answer 49

Thanks (• ◡•)|