Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

tkachenko Group Title

i need a little help

  • one year ago
  • one year ago

  • This Question is Open
  1. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Use a graphing calculator to solve the equation in the interval from 0 to 2. Round to the nearest hundredth. 7 cos 2t = 3

    • one year ago
  2. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i need it step by step to understand it

    • one year ago
  3. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    just type in 7* cos (2x) and where the graph crosses 3 is your answer. Make sure you in radians. Or degrees depends on what your q asks.

    • one year ago
  4. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    here it is when i pasted it to google.

    • one year ago
  5. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i got 6.995735789x

    • one year ago
  6. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    .64 i suppose.

    • one year ago
  7. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    idk about that with google i got .64

    • one year ago
  8. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    what about this one The water level varies from 12 inches at low tide to 52 inches at high tide. Low tide occurs at 9:15 a.m. and high tide occurs at 3:30 p.m. What is a cosine function that models the variation in inches above and below the water level as a function of time in hours since 9:15 a.m.?

    • one year ago
  9. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    use function manipulation

    • one year ago
  10. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    step bu step if you could

    • one year ago
  11. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    ok lets see if we can do this start with the graph of sin x |dw:1356583876231:dw| Lets see if we can manipluate it.

    • one year ago
  12. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    the rules are a*sin (x) a streches it up and down 1/a sin (x) shrinks it. a+sin(x) shifts it up -a+sin(x) shifts it down sin(x+a) shifts to the left TO the LEFT now lets combine those rules.

    • one year ago
  13. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so whats next then

    • one year ago
  14. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    1---- if you with me so far lets start The water level varies from 12 inches at low tide to 52 inches at high tide 52-12= 40 this gives us our amplitude which is difference so we get this 40*sin(x) http://tinyurl.com/dyqaluu <-graph but we want half of that cuzz of this |dw:1356585003441:dw| so now we have 20*sin(x) 2---- now we shift it up before i shift it up are you with me so far? there are a few more steps. shifting up and shifting left.

    • one year ago
  15. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    2-- shifting up so far we have this http://tinyurl.com/dyqaluu 20*sin(x) If you look at it the low point aka low tide is at -20 put we want it at +12 so we have shift up by an unknown amount we set up this equation -20+x=12 in my mind i read it as add x to -20 till you get 12 x=32 so now we have 20*sin(x)+32 with me so far. notice it getting more and more looking like our wanted graph now we shift left and right. With me so far? http://tinyurl.com/cpcmz73 graph

    • one year ago
  16. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yup with you

    • one year ago
  17. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    according to your info 3:30 is high tide. we can call that 3.5

    • one year ago
  18. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so whats the next step to this problem

    • one year ago
  19. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    Using 20*sin(x)+32 we have to get the hide tide to occur at 3.5 we shift it 20*sin(x-a)+32

    • one year ago
  20. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    ooh i noticed i messed up a bit by using sin. Using cos will be easier cuzz the amplitude is right at 0 hour |dw:1356585955086:dw|

    • one year ago
  21. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so should i erase sin and use cos

    • one year ago
  22. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    so just shift to right 3.5 so final answer is 20*cos(x-3.5)+32 ^ remember adding actually shifts to the left and subtracting shift right :)

    • one year ago
  23. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    yes use cos. Cos is nothing but sin shifted a certain amount to the left or right so that the coordinant of is 0,1 not 0,0

    • one year ago
  24. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so 20*cos(x-3.5)+32 is the final answer

    • one year ago
  25. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    yup did you understand? :)

    • one year ago
  26. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes thank you very much for all that help

    • one year ago
  27. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    let me know if it is correct

    • one year ago
  28. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

    • one year ago
  29. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    fan me so you dont lose touch. I am really interested in this problem. Have not done it in years. I am 99.99999999% sure it is right but since i have not done it in a while. there is still that .0000000000001% possibility i am rusty :D

    • one year ago
  30. tkachenko Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    its okay

    • one year ago
  31. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    there is also the odds that i need to stretch it horisontally

    • one year ago
  32. timo86m Group Title
    Best Response
    You've already chosen the best response.
    Medals 6

    oh yup you do have to stretch |dw:1356587327217:dw|

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.