simplify 6√128
128 / 4 = 36 36/ 4 = 8 8/4 = 2 therefore \[\sqrt{128} = \sqrt{4} \sqrt{4} \sqrt{4} \sqrt{2}\] = (2) (2) (2) \[\sqrt{2}\]
sorry the squareroot 2 is supposed to be next to the rest of the 2s
as a result, \[6\sqrt{128} = 6 \times 8\sqrt{2} = 48\sqrt{2}\]
Thanks a lot nali
u r welcome
How about √29. I answered. Not a square
what is the question?
Simplify √29
the trick is to look for perfect squares such as 4, 9, 25, 36, 49
So 29 is not a perfect square
yes just like 128
but u could divide 29 into a perfect square and another number
do u get it?
No
Still. Confusing
I answered. Simply the square root of 29 in my homework
i divided 128 by 4 becuz it was a perfect square then i kept dividing by 4 untill i got \[\sqrt{2}\] then i had to stop and left it as 48\[\sqrt{2}\]
wait iam about to answer ur question
Thanks nali. I understand about 128 cause it is perfect squate but 29 is not
\[29 can be divided by 4 and answer will be 7.25\]
sorry about that
29 can be devided by 4 and the answer will be 7.25
so \[\sqrt{29}= \sqrt{4} \times \sqrt{7.25}\]
= \[2\sqrt{7.25}\]
Thanks alot
u r welcome
Find consecutive integers. M and n such that given number is between m and n or state that the given number is not a real number. 1. -√115
sorry i can not help you in this one u should copy this question in the ask a question box so more people can help u
How about this. Find two fractions. Between. 2/3 and 3/4
I got more than 500 homework. And I don't. Know how to answer 5 of the question. I'm sorry if I bother u a lot
Do I need to devide it by 2
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