I am working on number 25, and and I got up to -10x to the -6 over y to the -2. I know that you would flip the x-6 down to become 1/x6. However, what do you do about the y-2?
Your mistake with #25 was in how you applied the -2 exponent. You apply it to EACH factor (5 and x^3). You were correct for the x^-6 part, but 5^-2 is NOT= -10... it is equal to 1/5^2, better known as 1/25.
okay...so I did number 24 because i know that it wasn't homework but i wanted to get practice.
the problem is-- 12(x^2y^4)^0/-6x^-2y^2
heres how i did it:
i took out the (x^2y^4)^0 becasue i wasn't sure but my brother said that anything to the zero power equals 1 (sry i forgot). so now i have: 12/-6x^-2y^2 then i put the x^-2 "up" so now i have: 12x^2/-6y^2 then i did some more symplifying and got:
-2x^2/y^2 <-- is that right?!?!?
i hope so! please someone try to get back to me asap...i have the test 2morro!! :(
7 comments:
I am working on number 25, and and I got up to -10x to the -6 over y to the -2. I know that you would flip the x-6 down to become 1/x6. However, what do you do about the y-2?
when doing number 21, why is the answer 4/b4 instead of 4 1/b4?
(For clarity, b^4 is b to the 4th power)
For pg 431, q21:
I think you are asking, is 4 "times" 1/b^4 a different answer than 4/b^4?
The answer is they are the same, however 4/b^4 would be considered simplest form.
Does that answer your question?
pg 431 q25
Two approaches (there are more)
Approach 1)
===========
Apply the -2 exponent first, yielding:
5^-2 x^-6 / y^-2
each factor "flips" yielding:
y^2 / 5^2 x^6 which= y^2 /25x^6
Approach 2
===========
Flip the whole schmegeggy first, yielding:
( y / 5x^3)^2
Then, apply the exponent to numerator and denominator, yielding:
y^2 / 5^2 x^6 then simplify to:
y^2 / 25x^6
Ca-peesh?
Your mistake with #25 was in how you applied the -2 exponent. You apply it to EACH factor (5 and x^3). You were correct for the x^-6 part, but 5^-2 is NOT= -10... it is equal to 1/5^2, better known as 1/25.
Yup or Nope?
okay...so I did number 24 because i know that it wasn't homework but i wanted to get practice.
the problem is--
12(x^2y^4)^0/-6x^-2y^2
heres how i did it:
i took out the (x^2y^4)^0 becasue i wasn't sure but my brother said that anything to the zero power equals 1 (sry i forgot).
so now i have:
12/-6x^-2y^2
then i put the x^-2 "up"
so now i have:
12x^2/-6y^2
then i did some more symplifying and got:
-2x^2/y^2 <-- is that right?!?!?
i hope so! please someone try to get back to me asap...i have the test 2morro!! :(
That is CORRECT, sir! (uh, madam)
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