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Math Help - Clueless

  1. #1
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    Clueless

    Can someone give me a hand with these? I'm at a total loss.

    View problems here.

    Then a word problem (these always throw me in a loop).

    Then a word problem too: "The height of a triangle is 4 inches less than the base. The area is 30 square inches. Find the height."
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  2. #2
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    Quote Originally Posted by SailorCardKnight View Post
    Can someone give me a hand with these? I'm at a total loss.

    View problems here.

    Then a word problem (these always throw me in a loop).

    Then a word problem too: "The height of a triangle is 4 inches less than the base. The area is 30 square inches. Find the height."
    For the word problem, note:

    The area of a triangle is 1/2 x length of the base x height.
    Call the length of the base "x". What is the height relative to this x?
    Then you can use the "1/2 x length of the base x height" to solve for x...
    Last edited by Unenlightened; November 24th 2009 at 06:48 AM. Reason: typo
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by SailorCardKnight View Post
    Can someone give me a hand with these? I'm at a total loss.

    View problems here.

    Then a word problem (these always throw me in a loop).

    Then a word problem too: "The height of a triangle is 4 inches less than the base. The area is 30 square inches. Find the height."
    Hi SailorCardKnight,

    \frac{a^2-16}{4a} \cdot \frac{8a}{a-4}

    First, factor the numerator of the first fraction:

    \frac{(a-4)(a+4)}{4a} \cdot \frac{8a}{a-4}

    Multiply, then simplify:

    \frac{(a-4)(a+4)(8a)}{(a-4)(4a)}

    \boxed{2(a+4)}

    Your second problem is a little difficult to read. I'll venture a guess:

    5-\frac{2}{5}x=\frac{7}{3}

    If this is not exactly it, you can still use it as a model.

    First, multiply by the least common multiple of the divisors to clear the fractions.

    15(5)-15\left(\frac{2}{5}x\right)=15\left(\frac{7}{3}\ri  ght)

    75-6x=35

    Subtract 75 from both sides and then divide by -6 and you're done.



    Now, onto your last one.

    y=-\frac{3}{4}x-17

    This is already in slope-intercept form, so we can easily tell that the slope is -\frac{3}{4}

    Now, you want the slope of a line that is perpendicular to this one, so we must use the negative reciprocal of this slope.

    The negative reciprocal of -\frac{3}{4} is \frac{4}{3}

    Using point (-1, 4) and our newly acquired slope of \frac{3}{4}, we can use the point slope form of the linear equation to find our required equation.

    y-y_1=m(x-x_1)

    It's just a matter of substitution now.
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  4. #4
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    You guessed right on that one problem masters, thats exactly how I have it wrote. (I will admit my handwriting is awful...)

    Thanks for the help you guys. ^^

    I'm so glad I found this forum, you all have been a big help so far, I owe ya all big.
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