# Math 9 Midterm Review

Question 1:

Chapter 3 review pg. 120 q 18 c

(2)6 (2)2 – 13 x 2  0 / (-1 + 2  2) 5

2 to the power of 6 and 2 to the power of 2 minus 13 times 2 to the power of 0 over negative 1 plus 2 to the power of 2 to the power of 5.

solve?

first rewrite 2 to the power of 0 as 1, and then rewrite terms that are multiplied 1 as the term. 2 to the power of 6 equals 64 so replace, also replace 2 to the power of 2 to get 4. Then multiply 64.4 to get 256 then minus 256 and 13 to get 243. 2 to the power of 2 equals 4 then add  -1 and 4 to get 3. 3 to the power of 5 equals 243 and now you have 243 over 243 and simplify to get 1.

https://www.freemathhelp.com/exponents.html

http://www.purplemath.com/modules/exponent.htm

Question 2:

Chapter 3 review pg. 120 q 18 d

(-10)10 + (-22)0 – ( 3 / 5 )2

negative 1 to the power of 10 plus negative 22 to the power of 0 minus 3 over 5 to the power of 2.

solve?

first, -22 to the power of 0 is the same as 1 so rewrite, then solve -1 to the power of 10 is 1. Group together any like terms, so you will have 1 + 1 and minus 3 / 5 to the power of 2. Add 1 +1 to get 2, then you have 2 – 3/5 to the power 2. Solve and then you will have 1 and 16 / 25.

https://www.mathsisfun.com/algebra/exponents-using.html

http://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx

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