80sqrt2+20sqrt
this square root addition adds up to 10sqrt5
2006-10-26 07:52:05 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
What are you solving for? I cant tell if you mean 80squareroot 2+ 20square root two or what the sqare root next to the 20 is.
Square root addition means that if you have like
80sqroot2+ 20sqroot2=100sqroot2
2006-10-26 07:58:08 · answer #1 · answered by asd589 2 · 0⤊ 1⤋
a million) The sq. root of one million+(24/25) needs to get replaced to (25/25)+(24/25). you are able to then upload both fractions to get 40 9/25. With sq. roots, it could be utilized to both the best and bottom of the fraction. so that you'll have the sq. root of 40 9 over the sq. root of 25 or 7/5. 2) an same theory contained in the first one might want to be utilized, getting the fraction to 50/26 or 25/13. then you definitely would take the sq. root of the best and bottom, getting 5/sqrt(13). To get rid of the sqrt(13) on the bottom, you multiply by technique of sqrt(13)/sqrt(13), it really is a twin of multiplying by technique of one. This receives you (5*sqrt(13))/13.
2016-12-05 06:25:55 · answer #2 · answered by louttit 4 · 0⤊ 0⤋
Sorry, but you have stated the problem incorrectly and it can not be solved as stated. The way it reads right now is:
80 times the squareroot of 2 + 20 the squareroot of ?
Please read problem again and then submit an accurate version of the problem.
2006-10-26 08:07:49 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
your question is actually
2sqrt80 + sqrt 20
= 2sqrt (16*5) + sqrt (4*5)
= 2*4 sqrt(5) + 2 sqrt(5)
= 8 sqrt5 + 2 sqrt5
= 10 sqrt5
2006-10-26 08:01:43 · answer #4 · answered by grandpa 4 · 1⤊ 0⤋
the question is not clear and the sum is NOT 10 rt5
unless your sum is 2 rt 80 +rt 20
2006-10-26 07:54:23 · answer #5 · answered by raj 7 · 0⤊ 0⤋
your question does not seem complete to me there should be another number after 20sqrt.
2006-10-26 08:28:55 · answer #6 · answered by wales4u_14 2 · 0⤊ 0⤋
it does. it does
2006-10-26 08:01:59 · answer #7 · answered by A 4 · 0⤊ 0⤋