Question 492650: I just took a math test and was disappointed to see that I got a "C." I had worked very hard and I DID feel confident in my abilities. Luckily, if I correct my test, I can have the opportunity to retake it and get up to an 80%. I simply need someone to show me how to do the problems I missed and explain what I did wrong.
Here are the questions:
2x^-4/(over)8x^-2 I got 4x^-2. I divided two into eight and had 4 leftover and then used the division of exponent rules (which I obviously used wrong). PLease explain. :)
#2
(3x^4y^-2/{over}4x^-5)^-2
I powered everything by the negative 2 first, and got 9x^-8y^-4/{over}16x^10, simplified and finally got 9/{over} 16x^2y^4
#3
1/4 + 3/8 = 3/4
I multiplied the whole thing by 4/1 and got 1+6y=3, used the addition method, divided by 6 and got y=1/3
#4
x-6 in absolute value lines is less than or equal to -8
I worked this out as a conjunction and got 14 is less than or equal to x is less than or equal to -2. I know the answer is no solution now, I just don't understand why. Could you please explain?
Thank you.
Answer by chessace(471)   (Show Source):
You can put this solution on YOUR website! 1. It's not enough to divide 2 into 8, you have to keep numerator and denominator straight. 1/4 not 4. Exponent OK unless they prefer 1/x^2 to x^-2.
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2. Similar; 3^-2 is not 9 but 1/9. (y^-2)^-2 is y^4, not y^-4.
4^-2 is 1/16 not 16.
Then x^-8/x^10 is 1/x^18 not 1/x^2.
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3. I assume (3/8)y [typo].
4 * (3/8) is 3/2, not 6.
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4. |x-6|<=-8
No solution is obvious right away, |anything|>=0, can't be < a negative.
Your work is OK, just another way of getting to No Solution.
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The good news is that if you are just more careful with the numerator v. denominator, 1 2 3 will go much better.
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