Question

@Notamathgenius

Answer 1

Yes?

Answer 2

P
E
M
D
A
S

Answer 3

ok but i dont know how to do the exponents

Answer 4

b^2 = B^2

Answer 5

A^0=0

Answer 6

but they want me to multiply

Answer 7

Yes do that on the inside of the () and then multiply the out come

Answer 8

but what about the one outside

Answer 9

actually A^0=1

Answer 10

but how do i work it out im confused

Answer 11

\[(A^2)^3=A^{2*3}\]

Answer 12

so its a^6

Answer 13

In that example yes

Answer 14

ok so the first part is 6A^2b^4

Answer 15

That's simplified already

Answer 16

???????

Answer 17

@goformit100

Answer 18

@uri

Answer 19

@ganeshie8

Answer 20

\(\large (3a^0b^2)(2a^{-3}b^2)^2\)

Answer 21

Is this the expression you have been working on ?

Answer 22

yees

Answer 23

ok, lets attack the second parenthesis first

Answer 24

\(\large (3a^0b^2)(2a^{-3}b^2)^2\)
^

Answer 25

square each term in that parenthesis

Answer 26

\(\large (3a^0b^2)(4a^{-6}b^4)\)

Answer 27

fine so far ?

Answer 28

shouldnt it be -4

Answer 29

nvm i see

Answer 30

:) good

Answer 31

next look at first parenthesis, the easy one -

Answer 32

we knw that anything to the power 0 = 1. so

Answer 33

\(\large (3b^2)(4a^{-6}b^4)\)

Answer 34

what happen to the a variable

Answer 35

it has a 0 above it, so \(a^0 = 1\)

Answer 36

if u have 0 in power, that variable just vanishes to 1

Answer 37

oo ok

Answer 38

somethingsomethingsomethin^0 = 1

Answer 39

next -

\(\large (3b^2)(4a^{-6}b^4)\)

multiply through

Answer 40

\(\large (3b^2)(4a^{-6}b^4)\)

\(\large (12b^6a^{-6})\)

Answer 41

see if that makes sense

Answer 42

yes

Answer 43

great ! its easy if u figure out by urself :)

Answer 44

next, the last step

Answer 45

we use this :-

\(\large x^{-m} = \frac{1}{x^m}\)

Answer 46

\(\large (12b^6a^{-6})\)

\(\large \frac{12b^6}{a^{6}}\)

Answer 47

oooo ok

Answer 48

:)