Question

[3c(4^2)-6c]/[5^2-4]
simplify the expression completely

Answer 1

\[\Large \frac{ 3c4^2-6c }{ 5^2-4 }\] right?

Answer 2

right

Answer 3

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. ;)

Answer 4

i can't understand
please the answer?

Answer 5

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.

Answer 6

can you draw me how to do it

Answer 7

please

Answer 8

What is 4^2?

What is 5^2?

Answer 9

=16
=25

Answer 10

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 = ?

Answer 11

c=16/3
c=5.3

Answer 12

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 = ?

Answer 13

=48

Answer 14

almost.... don't forget your c! So you get 48c for that term.

Now we have:

Answer 15

48c-6c

Answer 16

\[\Large \frac{ 48c-6c }{ 25-4 }\]

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).

Answer 17

42c/21

Answer 18

Excellent! And now you can reduce..... to end up with ?

Answer 19

14c/7

Answer 20

ok... and one more - you can still reduce further, you still have a common factor. :)

Answer 21

i dont know

Answer 22

Hmmm...... well, 14 = ? * 7 ??

Answer 23

2*7=14

Answer 24

Right, so 14/7 = ?

Answer 25

It's not fully simplified until you have cancelled all common factors.... so far you have:

\(\Large \dfrac{ 48c }{ 21} =\dfrac{ 14c }{ 7} =\dfrac{ ?c }{ ?} \)

Answer 26

7/7

Answer 27

oops, that should be 42c/21

Answer 28

14/7 = 7/7 ??

Answer 29

does 14 = 7 ?? :) C'mon, you know how to reduce this....

Answer 30

\(\Large \dfrac{ 42c }{ 21} =\dfrac{ 14c }{ 7} =\dfrac{ ?c }{ ?} \)

Answer 31

7c/7

Answer 32

\(\Large \dfrac{ 14 }{ 7} \neq\dfrac{ 7 }{ 7} \)

Answer 33

You told me that 2*7 = 14

So
\(\Large \dfrac{ 14c }{ 7} =\dfrac{7\cdot2c }{ 7} \)

right??

Answer 34

i dont know what to do

Answer 35

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.)

Answer 36

i got it

Answer 37

no
wait

Answer 38

im 9th
its algebra 1

Answer 39

\(\Large \dfrac{ 14c }{ 7} =\dfrac{7\cdot2c }{ 7}=\dfrac{7}{ 7}\cdot2c\)

Answer 40

\(\Large =\dfrac{\cancel7}{ \cancel7}\cdot2c=1\cdot2c\)

Answer 41

1.2c is the answer?

Answer 42

or 1*2c

Answer 43

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.

\(\Large =\dfrac{\cancel7}{ \cancel7}\cdot2c=1\cdot2c=2c\)

Answer 44

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.

Answer 45

thanks

Answer 46

so much thanks

Answer 47

You're welcome. :) happy to help!