Question

1.)−2xy^2z^3(2xy+3xy−4yx)

Answer 1

question?

Answer 2

perform the indicated operation..

Answer 3

Oh. LOL

Answer 4

-4x^2y^3z^3 -6x^2y^3z^3 +8xy^3z^4

Answer 5

I just expanded it.

Answer 6

im not going to multiply it?

Answer 7

You're just going to distribute the term to the 3 terms inside the parenthesis.

Answer 8

oh.. thank you..

Answer 9

can you help me again?

Answer 10

how about this?
2.) 2abc{abc[abc(abc)^2]^2}^2

Answer 11

2A^15b^15c^15

Answer 12

how did u get that?

Answer 13

first, solve the inner parenthesis.

Answer 14

it will become...

Answer 15

that makes 2abc{abc[abc(a^2b^2c^2]^2}^2

Answer 16

\[a^2b^2c^2\] ?

Answer 17

yeah.

Answer 18

then,?

Answer 19

2abc { abc [ abc(a^2b^2c^2) ]^2} ^2

then multiply "abc(a^2b^2c^2)" which makes a^3b^3c^3

Answer 20

which makes 2abc { abc [ a^3b^3c^3 ]^2} ^2

Answer 21

i get that part.. then,?

Answer 22

then do this : { abc [ a^3b^3c^3 ]^2}

Answer 23

you'll get {abc [a^6b^6c^6] }

Answer 24

which makes 2abc { abc [a^6b^6c^6] } ^2

Answer 25

then do this: { abc [a^6b^6c^6] }, you'll get a^7b^7c^7

Answer 26

which makes 2abc { a^7b^7c^7 } ^2

Answer 27

ok, i now i know y i get wrong.. i simplify [a^3b^3c^3] as \[a^9b^9c^9\]
hahahaha.. thats why im wrong...

Answer 28

Hahah. You divided the exponents by it's gcf? LOLOLOL

Answer 29

then you'll get 2abc { a^14b^14c^14 }

Answer 30

then muliply and get 2A^15b^15c^15 :]]

Answer 31

no, i didn't divide the exponents as gcf.. i thought \[(a^3b^3c^3)^2\]
is equal to \[a^9b^9c^9\]

Answer 32

hahahah.. my mistake, thank you again..

Answer 33

Oh. I thought 9 was 0. hahah no prob. :]

Answer 34

one more pls:)

Answer 35

Okay. Bring it on!

Answer 36

3.) -[a^2b^2c^2(abc)^2]^2. [a^2b^2c^2(abc)^2]^2

Answer 37

☺

Answer 38

\[-[a^2b^2c^2(abc)^2]^2 [a^2b^2c^2(abc)^2]^2\]
\[-[a^2b^2c^2(a^2b^2c^2)]^2 [a^2b^2c^2(a^2b^2c^2)]^2\]
\[-[a^4b^4c^4)]^2 [a^4b^4c^4]^2\]
\[-[a^8b^8c^8] [a^8b^8c^8]\]
\[-[a^{16}b^{16}c^{16}]\]

Answer 39

my mistake again, in simplifying the exponent.. hehehehe:D hate my self..

Answer 40

thank you so much...

Answer 41

No problem :DD

Answer 42

do i need to distribute the - sign?

Answer 43

no need. that is only one term.

Answer 44

I have to go now, thanks for the problems. LOL

Answer 45

ok.. you are such a math wizard:))

Answer 46

ok..thank you

Answer 47

Hahah thanks.