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Math Help - A few questions about INEQUALITIES

  1. #1
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    A few questions about INEQUALITIES

    Okay, in the first one...it says solve for (x^2+2x) / (x^2-8) < 0
    So I did everything and solves for it, but then I had to write in which intervals the solutions were...so I PUT [-3, -2] U [0,3) but for some reason the correct answer says that it's (-3,-2) with parentheses before the 3 instead of brackets...and I have NO idea why...

    The same question for

    square root( x / x^2-2x-35) > 0..I solved, and found the intervals to be [-5,0)U[7, infinity) but the correct answer is (-5, 0] with parentheses instead of brackets before the 5, and brackets instead of parentheses after the o, and U(7, infinity), with parentheses before the 7 instead of brackets...

    I don't get when it's parentheses or brackets

    Finally,

    a word problem...

    A projectile is fired straight upward from ground level with an initial velocity of 160 feet per second.

    a. At what instant will it be back at ground level?
    b. When will the height exceed 384 feet?

    I know I can find b if I first write an inequality for what happened, but..I have no idea how...and the book says the equation should be:

    -16t^2 + 160t = 0...where in the world did that -16t^2 come from???

    Appreciate any help, thank you!
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  2. #2
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    Quote Originally Posted by jonjon1324 View Post
    Okay, in the first one...it says solve for (x^2+2x) / (x^2-8) < 0
    Is the denominator supposed to be x^2- 9? If not, where did you get the "3" and "-3"?

    So I did everything and solves for it, but then I had to write in which intervals the solutions were...so I PUT [-3, -2] U [0,3) but for some reason the correct answer says that it's (-3,-2) with parentheses before the 3 instead of brackets...and I have NO idea why...
    Assuming the denominator is actually [tex]x^2- 9[/itex], then the left sided is not defined for x= 3 or -3 so they cannot be included in the solution set. If that is really " \le" and not "<" then the solution set is (-3, -2]\cup [0, 3).

    The same question for

    square root( x / x^2-2x-35) > 0..I solved, and found the intervals to be [-5,0)U[7, infinity) but the correct answer is (-5, 0] with parentheses instead of brackets before the 5, and brackets instead of parentheses after the o, and U(7, infinity), with parentheses before the 7 instead of brackets...

    I don't get when it's parentheses or brackets
    The denominator, x^2- 2x- 35= (x- 7)(x+ 5) is 0 at x= -5 and 7 so the left side is undefined. x= -5 and x= 7 cannot be included in the solution set.

    I presume you understand that [-5, 0) means " -5\le x< 0 while (-5, 0] means " -5< x\le 0".

    Finally,

    a word problem...

    A projectile is fired straight upward from ground level with an initial velocity of 160 feet per second.

    a. At what instant will it be back at ground level?
    b. When will the height exceed 384 feet?

    I know I can find b if I first write an inequality for what happened, but..I have no idea how...and the book says the equation should be:

    -16t^2 + 160t = 0...where in the world did that -16t^2 come from???

    Appreciate any help, thank you!
    The acceleration due to gravity is -32 feet per second per second. From that one can calculate that the speed after t seconds will be -32t+ <br />
v_0, where v_0 is the initial speed, and that the height after t seconds will be -16t^2+ v_0t+ h_0, where h_0 is the initial height.
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