May 24, 2019 / By Jewell
Question: I have 20 questions about exponents such as (2^2)^5 said as 2 to the 2nd power then to the fifth power. I need help desperately! Please my math teacher doesn't explain good enough!

Francine | 5 days ago
if it is (2^2)^5 simply multiply the inner exponent by the outer exponent, so 2 * 5 do this for all the problems.
👍 218 | 👎 5
Come on, these are so easy, just hit Start in the lower left, programs, accessories, and pick out the calculator. A percentage is entered into a calculator with decimals, two decimal places to the left. 25% would be .25 14 - multiply 65 by .45 15 - multiply 21 by .75 16 - Divide 60 by 85 17 - Multiply 75 by .70 (multiple ways to do this, you could also multiply by 30%, but you'd have to subtract that number from 75) 18 - Multiply 500 by 1.03 (if you multiply by just 3%, you still have to add that amount to 500, that's your interest) 19 - Similar as above, multiply 150 by 1.20 29 - This time they only want interest, so just take 450 multiplied by .02, and that's your interest

Darby
kk so for the 1st one million/2 dividing issues, you may in simple terms subtract the powers... so for the 1st difficulty, the respond ought to be 2 to the third potential (2^3) because of the fact the two^5 / 2^2= 2^3 (because of the fact 5-2 = 3) on each and every occasion the backside selection (which interior the 1st question, is two) is the the comparable on the two the numerator and the denominator, then u consistently subtract the powers and keep the backside selection the comparable. and for the 5th question, the respond may well be 10 to the third potential (10^3) because of the fact 10 is comparable to 10 to the 1st potential (10^one million) for the 2nd section, the 1st difficulty, (x^3)^2 the respond may well be x^6 (x to the 6th potential) because of the fact whenever you have the parentheses around a selection with an exponent raised to a selection, then you definately consistently multiply the two exponents and depart the backside selection (which interior the 1st difficulty is x) the comparable for the 2nd difficulty, it would be a^8 because of the fact multiplying the two exponents (4 and a couple of) equals 8 for the 5th difficulty, (z^3)^2 x (z^4)^2, the technique may well be: z^6 x z^8 = z^ 14 subsequently, the two exponents would be further because of the fact the exponents are not in a sort like this : (x^6)^3 <---- in those situations you frequently multiply the exponents in spite of the shown fact that, in situations like this : a^2 x a^3 <---- in those situations you frequently upload the exponents yet then there r the situations the place the backside numbers are not the comparable, yet it quite is yet another tale yet ur issues all have the comparable base numbers fro each and each difficulty so as that heavily isn't an argument :) desire that helped!!
👍 90 | 👎 -2

Braidy
(2^2)^5 Always perform operations inside the parenthesis first so 2^2 (two squared) is four. Next you have 4^5, four to the fifth power which is 4x4x4x4x4 and that is 1,024 so (2^2)^5 = 1,024 .
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Albina
2, first, is multiplied by itself twice: ? 2 * 2 = 4 ... then the answer of that will be multiplied by itself five times: ? 4 * 4 * 4 * 4 * 4 = 1 024 =)(=
👍 82 | 👎 -16