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Math HW

  1. Aug 19, 2008 #1
    1. Find all ordered pairs (x, y) that satisfy the system of equations given below:
    * (x + 1)^2 + |(x + 1)(y − 2)| + (y − 2)^2 = 7
    * x^2 + y^2 + 2x − 4y − 5 = 0

    2. If the area of a certain rectangle is 48, and the length of its diagonal is 10, find its perimeter.

    here's what i did.

    1. 2nd equation is equal to "(x+1)^2+(y-2)^2=10"
    what should i do next?

    2. i know that the answer to this is 28 because the sides are equal to 6 & 8... but how do i solve for it?
     
  2. jcsd
  3. Aug 19, 2008 #2

    statdad

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    Homework Helper

    The second equation is the 'standard form' of a shape you should be familiar with at this time.

    I'm not sure how you found the answer to 3, since you don't have any work to present (trial and error?)

    Try drawing a rectangle, one side labeled [tex] x [/tex], the other [tex] y [/tex]. Since the area is [tex] 48 [/tex], you know that [tex] xy = 48 [/tex] so

    [tex]
    y = \frac{48} x
    [/tex]

    Now substitute into the formula for finding the known length of the diagonal and solve.

    For your first question, I have a question. Did you intend to type

    [tex]
    (x+1)^2 + (x+1)(y-2) + (y-2)^2 = 7
    [/tex]

    or was it what I have below, with the middle term in absolute values? (this is what seems to appear)

    [tex]
    (x+1)^2 + |(x+1)(y-2)| + (y-2)^2 = 7
    [/tex]

    I'm guessing there aren't any absolute value signs in the original problem.
     
  4. Aug 19, 2008 #3
    actually, there are absolute value signs.
     
  5. Aug 19, 2008 #4
    so... in the 2nd question.

    (48/x)^2 + x^2 = 100

    2304/(w^2) + (w^2) = 100

    [2304 + (w^4)]/ (w^2) = 100

    what should i do to get rid of those exponents?
     
  6. Aug 19, 2008 #5

    HallsofIvy

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    From the second equation you know, as you say, that (x+1)2+ (y-2)2= 10. You can replace those same terms in the first equation by 10:
    (x+1)2+ |(x+1)(y-2)|+ (y-2)2= 7 becomes
    |(x+1)(y-2)|+ 10= 7 or |(x+1)(y-2)|= -3. Now, what values of x and y satify that?


    If the sides have length x and y, then you know that xy= 48 and x2+ y2= 10. You need to solve those two equations. From the first, y= 48/x so the second becomes x2+ 482/x2= 10. Multiply both sides of the equation by x2 and you get a quadratic equation for x2.
     
  7. Aug 19, 2008 #6

    HallsofIvy

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    Multiply both sides of the equation to get 2304+ w4= 100w2 which is the same as (w2)2- 100(w2)+ 2304= 0, a quadratic equation in w2.
     
  8. Aug 19, 2008 #7
    none? coz absolute values can nver be negative,ryt?

    so x^2 = 64

    x= 8

    y = 6?
     
  9. Aug 19, 2008 #8
    3. Find the domain of the following functions:
    (a) f(x) =
    (x^2 − 1)^(1/3) − 1 / [|3-x| - (x-3)^(1/3) - 3]

    (b) f(x) =
    (x^2 − 2x + 1)^(1/2) / [|2x − 3| + 1]

    (c) f(x) = x^3 / (|3 − x| − (x – 3)^(1/3))

    how do i do these?
     
  10. Aug 19, 2008 #9

    statdad

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    The domain of a rational function depends only on the denominator, as long as there are no radicals or even roots in the numerator.
    The domain for each can be found by determining all numbers (if any) that make the denominator zero: once those values are eliminated, the domain is all values that remain.
     
  11. Aug 19, 2008 #10
    for a. x>12

    b. all real #s

    c.how do i do this?
     
  12. Aug 19, 2008 #11

    HallsofIvy

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    What values of x make |3- x|- (x-3)1/3= 0?
     
  13. Aug 19, 2008 #12
    is the answer 3? how about the other numbers, are they correct?
     
  14. Aug 20, 2008 #13

    HallsofIvy

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    I asked a question you are supposed to ANSWER, not ask another question!

    |3-x|- (x-3)1/3= 0 is the same as |x- 3|= (x-3)1/3.

    Now, suppose x> 3. then |x-3|= x- 3= (x-3)1/3. Can you solve that?

    If x< 3, then |x-3|= 3- x= (x-3)1/3.
     
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