Hard thinking questions

  1. 1. The problem statement, all variables and given/known data

    a) given, a + 5 and 2a-1, one is 40% greater than the other one, solve for a.

    b) given, 5x[tex]^{4}[/tex] + 4x[tex]^{3}[/tex] + 3x[tex]^{2}[/tex] + Px + Q. When divided by x[tex]^{2}[/tex] - 1, you get a remainder of 0, solve for p and q.

    c) given csc(6b + [tex]\frac{pi}{8}[/tex]) = sec(2b - [tex]\frac{pi}{8}[/tex]), solve for b.

    2. Relevant equations


    3. The attempt at a solution

    please help, i can't get started on these three at all... :confused:

    edit:

    c) i know that cscx = 1/sinx and secx = 1/cosx, i then cross multiplied to get [tex]cos(2b-pi/8) = sin(6b+pi/8)...
     
    Last edited: Oct 21, 2010
  2. jcsd
  3. Question a) shouldn't be a problem at all.
    Any attempt by you ?
     
  4. Would it be (a+5) = .4(2a-1) + 2a-1 ? or .4(a+5) + (a+5) = 2a - 1 ?
     
  5. It's not clear from the text. Choose one and go on, or solve both in sequence.
     
  6. well then you get a+5 = 0.8a - 0.4 + 2a -1
    a+5 = 2.8a - 1.4
    1.8a - 6.4 = 0
    1.8a = 6.4
    a = 3.55555556 :confused:

    or

    2a - 1 = .4a + 2 + a + 5
    2a - 1 = 1.4a + 7
    0.6a - 8 = 0
    0.6a = 8
    a = 13.3333333

    therefore a can be 13.333 or 3.555 ? :confused:
     
  7. yes.
     
  8. thanks :smile: could you help me get started on b) ?

    Thanks for your time!
     
  9. b)
    Make this multiplication [tex] (x^2-1)(Ax^2+Bx+C)[/tex]
     
  10. that would give you [tex] ax^4 + bx^3 + (c - a)x^2 - bx - c[/tex]. Is that what you mean?
     
  11. Ok, so now give the solution. It's really easy.
     
  12. do you use the original equation? so a = 5, b = 4, and c = 3?

    but that doesn't work... cause then you get:[tex]5x^4 + 4x^3 + (3 - 5)x^2 - 4x - 3[/tex]
     
  13. You realize by yourself there's something wrong. Ok.

    You should come alone to the solution. The next step would be giving you the solution.
     
  14. I don't see what I could possibly do to get the answer :confused: I know that c - a = 3, that's all... but c itself is 3 :confused:
     
  15. Why you say "c itself is 3"... ???
    Tell me where is written that ?
     
  16. oh, well i based it on the original... a = 5, b = 4, c = 3...

    well, then if that's the case,

    [tex]ax^4 + bx^3 + (c - a)x^2 - bx - c[/tex]
    [tex]=5x^4 + 4x^3 + (c - 5)x^2 - 4x - c[/tex]
    therefore
    [tex]
    c = 8 [/tex]?
    and therefore:
    [tex]P = -4[/tex]
    [tex]Q = -8[/tex]

    is that what you meant?

    Thanks for your help! Means a lot!
     
    Last edited: Oct 21, 2010
  17. Not quite....
    You solve for the cases where one term is 40% of the other, not 40% greater then the other, as the question asked!

    In other words, suppose you have a number (let's say 20) and you want to find the value that is 40% greater.
    0.4 * 20 = 8 which is clearly not 40% greater. In fact, it is less!

    What would you do to determine the value that is 40% greater than 20?
     
  18. Well you would do .4 * 20 + 20, a value 40% greater then 20 is 28?
     
  19. Yes, but that simplifies...

    0.4 * 20 + 20 = (0.4 + 1) * 20 = 1.4 * 20

    Therefore, a number that is 40% larger than x is 1.4 * x
     
  20. well at first glance (and i do mean i only "glanced" at it for a split second), i thought the same thing...here are the equations he set up:
    look closer at the right-hand side of the first equation. 0.4(2a-1) + 2a-1 = 1.4(2a-1), and the original equation becomes a+5 = 1.4(2a-1), which certainly implies that a+5 is 40% larger than 2a-1. likewise, we can see that the left-hand side of the 2nd equation, 0.4(a+5) + (a+5), equals 1.4(a+5). and so equation 2 becomes 1.4(a+5) = 2a-1, which certainly implies that 2a-1 is 40% greater than a+5.


    your calculations are correct - you just chose to represent the quantity on one side ofthe equation as 0.4x + x instead of 1.4x.
     
  21. tiny-tim

    tiny-tim 26,041
    Science Advisor
    Homework Helper

    Hi I Like Pi!! :smile:

    (since you like them so much, have a pi: π :wink:)
    Yes, and now use cosθ = sin(π/2 - θ) to get that in the form sinA = sinB. :smile:
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?