This week in Math 10 we learned about negative integers, and how to turn them into positive.

In the image we are multiplying (-7 x^3 x^-5)(x^2 x^-3) first I used the power of multiplication to make X all one letter with one exponent, so we add the exponents over all the X’s to get -3. Now we need to make all these numbers positive including the 7, so I divide by -1 to make the 7 positive and the -3 exponent over X positive. The 7 goes on top because the -1 goes invisible so I had to put a number up there, and because a -#/-# equals a positive number we can put the 7 on top, because it has no visible negative exponent.

In the image we see two different bases for the exponent, meaning we can not multiply or divide any of our exponents, the only thing we can do here is make the M^-5 and M^5. To do that we simply divide m^-5/1 to make it 1 over m^5. now n^2 is really n^2/1 so we can multiply n^2 * 1 to get the n^2 on top of our fraction. And of course there is a 1 under our n^2 so we can multiply 1 * m^5, which will stay the same. Thus resulting in our answer being n^2/M^5.

In the image {-7^2 * 8^-2} we can solve the answer by simply doing {-7 * -7} which equals 49 because a negative * a negative = a positive. for the 8^-2 the negative exponent tells us that it is on the wrong side of the fraction which is invisible so we divide one by 8^-2 to get {1/8^-2} and we can multiply that by {49/1} to get {49/8^2} then we can simply do 8^2 which will get you 64.