Positive exponent help!!!!!?

Question

ok my teacher just handed this paper to me and told me to do it, im doing my best and i only have two left will you please help me and tell me how you got it thanks!!!

(5Y^2)/(45y^7)

and

20x^5 * (4x^2)^-2

please tell me step by step thanks so much!!!

Answer 1

I'm going to assume that you need entirely positive exponents, given the title...

You need to reduce these?  Okay...

For the first, (5y^2)/(45y^7)

We can reduce the exponents; dividing exponents means we can subtract them (a^x)/(a^y) = a^(x-y)

We can get rid of the exponent two on top if we subtract two from the bottom... that gets rid of the top exponent!  (5* 1)/(45y^5)

The 1 represents y^0; any number to the zero power is 1.

The coefficients reduce, since both are divisible by five... 5/45 = 1/9

Leaving us with (1)/(9y^5)

:)

Let's press on!

20x^5 * (4x^2)^-2

Negative exponents are the equivalent of dividing by a positive exponent... (a^-x) = (1)/(a^x)  In our case, the second bit is all being raised to a negative power, and the first bit replaces the 1 in the example there.

Doing this, we get (20x^5)/((4x^2)^2)

Let's take care of the lower bit.  Raising a power to a power multiplies them.  Example: (a^x)^y = a^(xy)

2*2 is 4, but the extra power out there also squares the exponent; we need to square the 4, too.  So our lower term is 16x^4.

Now we have (20x^5)/(16x^4)

We know we can reduce fractions by subtracting them, which means that we can eliminate the lower exponent entirely.  5-4 = 1, so...

(20x^1)/(16)

the 20 and 16 reduce, both are divisible by 4, so we finally get (5x)/(4)

Hope that makes sense (^_^;; )