anonymous
  • anonymous
[3c(4^2)-6c]/[5^2-4] simplify the expression completely
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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DebbieG
  • DebbieG
\[\Large \frac{ 3c4^2-6c }{ 5^2-4 }\] right?
anonymous
  • anonymous
right
DebbieG
  • DebbieG
So... apply PEMDAS to the num'r and den'r (separately). Evaluate the exponents, then you'll have some multiplication in the num'r (the 3 * 4^2), then add/subtract the like terms. You'll finally get to where you have a single term (a monomial) in the num'r, and just a number in the den'r. You might be able to reduce at that point. ;)

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anonymous
  • anonymous
i can't understand please the answer?
DebbieG
  • DebbieG
As is the site policy, I will not give you the answer. http://openstudy.com/code-of-conduct Anyway, it is of no use to you to have the answer without understanding how to do the problem, as clearly your teacher thinks you should be able to do. If you want me to help you UNDERSTAND HOW to do it, I will. If you don't want to understand it, then I won't bother. Up to you.
anonymous
  • anonymous
can you draw me how to do it
anonymous
  • anonymous
please
DebbieG
  • DebbieG
What is 4^2? What is 5^2?
anonymous
  • anonymous
=16 =25
DebbieG
  • DebbieG
Ok good... so now we have \[\Large \frac{ 3c(16)-6c }{ 25-4 }\] Now do the product in the first term of the num'r: 3*c*16 = ?
anonymous
  • anonymous
c=16/3 c=5.3
DebbieG
  • DebbieG
Ack... no. First of all, we aren't SOLVING for c. it's a TERM not an EQUATION. Just simplify the product: 3*c*16 = ? E.g., 2*m*12 = 24m 5*z*(-7)=-35z so what is 3*c*16 = ?
anonymous
  • anonymous
=48
DebbieG
  • DebbieG
almost.... don't forget your c! So you get 48c for that term. Now we have:
anonymous
  • anonymous
48c-6c
DebbieG
  • DebbieG
\[\Large \frac{ 48c-6c }{ 25-4 }\] OK, now one at a time we'll do the subtraction. In the den'r, you have 25 - 4 =? In the num'r, you have 48c - 6c =? (remember that you do this kind of addition/subtraction just by adding/subtracting the coefficients).
anonymous
  • anonymous
42c/21
DebbieG
  • DebbieG
Excellent! And now you can reduce..... to end up with ?
anonymous
  • anonymous
14c/7
DebbieG
  • DebbieG
ok... and one more - you can still reduce further, you still have a common factor. :)
anonymous
  • anonymous
i dont know
DebbieG
  • DebbieG
Hmmm...... well, 14 = ? * 7 ??
anonymous
  • anonymous
2*7=14
DebbieG
  • DebbieG
Right, so 14/7 = ?
DebbieG
  • DebbieG
It's not fully simplified until you have cancelled all common factors.... so far you have: \(\Large \dfrac{ 48c }{ 21} =\dfrac{ 14c }{ 7} =\dfrac{ ?c }{ ?} \)
anonymous
  • anonymous
7/7
DebbieG
  • DebbieG
oops, that should be 42c/21
DebbieG
  • DebbieG
14/7 = 7/7 ??
DebbieG
  • DebbieG
does 14 = 7 ?? :) C'mon, you know how to reduce this....
DebbieG
  • DebbieG
\(\Large \dfrac{ 42c }{ 21} =\dfrac{ 14c }{ 7} =\dfrac{ ?c }{ ?} \)
anonymous
  • anonymous
7c/7
DebbieG
  • DebbieG
\(\Large \dfrac{ 14 }{ 7} \neq\dfrac{ 7 }{ 7} \)
DebbieG
  • DebbieG
You told me that 2*7 = 14 So \(\Large \dfrac{ 14c }{ 7} =\dfrac{7\cdot2c }{ 7} \) right??
anonymous
  • anonymous
i dont know what to do
DebbieG
  • DebbieG
If you don't mind my asking, what class is this for? What grade are you in? (I'm just wondering... it helps me understand what your math knowledge is.)
anonymous
  • anonymous
i got it
anonymous
  • anonymous
no wait
anonymous
  • anonymous
im 9th its algebra 1
DebbieG
  • DebbieG
\(\Large \dfrac{ 14c }{ 7} =\dfrac{7\cdot2c }{ 7}=\dfrac{7}{ 7}\cdot2c\)
DebbieG
  • DebbieG
\(\Large =\dfrac{\cancel7}{ \cancel7}\cdot2c=1\cdot2c\)
anonymous
  • anonymous
1.2c is the answer?
anonymous
  • anonymous
or 1*2c
DebbieG
  • DebbieG
OK, you should probably review your multiplication tables and basic products, factors, etc. it isn't 1.2c.... it's 1 TIMES 2c. You don't need the 1, it was just to show you why the factors reduce as they do. \(\Large =\dfrac{\cancel7}{ \cancel7}\cdot2c=1\cdot2c=2c\)
DebbieG
  • DebbieG
But it is VERY IMPORTANT that you UNDERSTAND how to do this. I walked you through it, but I can't help you on your test. :) YOU need to grasp it.
anonymous
  • anonymous
thanks
anonymous
  • anonymous
so much thanks
DebbieG
  • DebbieG
You're welcome. :) happy to help!

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