SOLUTION: 14-3(2³-4x2)/10+2[(-3)²-3-(-2)²] My book says the answer is 1, but I'm having trouble arriving at that number. Could someone please show me their step-by-step solution to this prob

Algebra ->  Signed-numbers -> SOLUTION: 14-3(2³-4x2)/10+2[(-3)²-3-(-2)²] My book says the answer is 1, but I'm having trouble arriving at that number. Could someone please show me their step-by-step solution to this prob      Log On


   



Question 795353: 14-3(2³-4x2)/10+2[(-3)²-3-(-2)²] My book says the answer is 1, but I'm having trouble arriving at that number. Could someone please show me their step-by-step solution to this problem. I would really appreciate it. Thankyou for all the work you guys do.
Found 3 solutions by Cromlix, stanbon, MathTherapy:
Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
14-3(2³-4x2)/10+2[(-3)²-3-(-2)²]
14 -3(8 - 8)/ 10 + 2[9 - 3 - 4}
14 - 3(0) / 10 + 2[2]
14 - 0/10 + 4
14/14
= 1
Hope this helps.
:-)

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
14-3(2³-4x2)/10+2[(-3)²-3-(-2)²]
--------------
= 14 - 3(8-4x^2)/10 + 2[9-3-(4)]
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= 14 -3(4)(2-x^2)/10 + 2[2]
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= 14 -(6/5)(2-x^2)+4
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Note: That certainly is not "1".
Please check your post.
Cheers,
Stan H.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

14-3(2³-4x2)/10+2[(-3)²-3-(-2)²] My book says the answer is 1, but I'm having trouble arriving at that number. Could someone please show me their step-by-step solution to this problem. I would really appreciate it. Thankyou for all the work you guys do.

As written, the answer's 18. In other words, if it's:

14+-+3%282%5E3+-+4+%2A+2%29%2F10+%2B+2%28%28-+3%29%5E2+-+3+-+%28+-+2%29%5E2%29, then it's 18. However, if:

%2814+-+3%282%5E3+-+4+%2A+2%29%29%2F%2810+%2B+2%28%28-+3%29%5E2+-+3+-+%28+-+2%29%5E2%29%29, then it's 1.

You can do the check!!

Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com. Further help is available, online or in-person, for a fee, obviously.

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