SOLUTION: divide and simplify so that there are no radicals in the denominator. show your steps.
1. square root 7 over 100
2. square root of 18 + square root 24 over square root of 6
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Problems-with-consecutive-odd-even-integers
-> SOLUTION: divide and simplify so that there are no radicals in the denominator. show your steps.
1. square root 7 over 100
2. square root of 18 + square root 24 over square root of 6
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Question 252926: divide and simplify so that there are no radicals in the denominator. show your steps.
1. square root 7 over 100
2. square root of 18 + square root 24 over square root of 6
3. 8 over square root of 6
You can put this solution on YOUR website! 1. no radical in the denominator
2.sqrt(18)+sqrt(24) /(sqrt(6))
factor
sqrt(6)*sqrt(3)+sqrt(6)*sqrt(4)/sqrt(6)
sqrt(6)*(sqrt(3)+sqrt(4))/sqrt(6)
I leave the rest for you
3.8/sqrt(6)
multiply by sqrt(6)/sqrt(6)
3.8*(sqrt(6)/6
I leave it for you to simplify.
You can put this solution on YOUR website! divide and simplify ao that there are no radicals in the denominator. show your steps.
1. square root 7 over 100
2. square root of 18 + square root 24 over square root of 6
3. 8 over square root of 6
You can put this solution on YOUR website! divide and simplify so that there are no radicals in the denominator. show your steps.
1. square root 7 over 100
I will assume the question looks like this:
we get
which is /
--
2. square root of 18 + square root 24 over square root of 6
( + ) /
simplify sqrt(18) as 3sqrt(2) and sqrt(24) as 4sqrt(6)
() /
multiply by conjugate, sqrt(6), on both numerator and denominator to get
()* /
distribute to get
() /
simplify the numerator to get /
and then factor out a 6 to get .
--
3. 8 over square root of 6
multiply by conjugate, sqrt(6) on both top and bottom to get
reduce to get
(
