1. Question 4

2. Question 4: Select all of the expressions with a value between 0 &1.
87 • 8-12 −3
( 1 3 ) 2 ∙ ( 1 3 ) 9
(−5) 6 (−5) 10 c
87+-12
= 8-5 =
=74+3
= 77
( 1 3 ) 2+9
=( 1 3 ) 11 c
There are two main things to stress in this problem. 1) they need to know their exponents rules. 2) they need to get all three answers correct to get credit for the problem.
For the answer to be correct, the students need to look for positive fractions, in order to be between 0 and 1.
For this solution, the students needed to remember that when the base is the same, you add the exponents together. So you add With that value, you get -5. So now you have 8 raised to the -5 power. You cannot have negative exponents. In order to make that exponent positive, you make it into a fraction and put 1 over the number. So you get 1/85. This stays a fraction when you multiply it out, so it is one of the answers.
For this solution, the students need to remember that when you divide fractions with the same base (big number), you subtract the exponents (little numbers). We get 7 raised to the power of 4 minus a negative three. The two negative signs in a row make it positive, so you get 7 to the power of 7. At this point, it is a not a fraction, but a whole number. Therefore, this is NOT a correct answer.
For this solution, the students need to recognize that the base is the same, so they can add the exponents together. The value of the exponents (2 + 9) gives you 11. So we have a fraction raised to the 11th power. This will remain a fraction when you multiply (1/3) 11 times.
For this solution the students will need to subtract the exponents, since the bases are the same. When you subtract 6 – 10, you get – 4. In order to make that exponent positive, you need to bring it down to the bottom of the fraction so you get 1/(-5)4. (-5)4 power ends up meaning you multiply -5 four times, which becomes a positive number. This is why it is an answer because it remains a fraction AND is positive.
Students should NOT be using a calculator for this problem. c
(-5)6 –10
=(-5)–4
= (-5)4