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
https://www.khanacademy.org/math/in-eighth-grade-math/exponents-powers-1/laws-exponents/v/exponent-rules-part-1
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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