find the square root of 1156 over 289
and square root of 0.000049
2007-04-16 20:20:32 · 4 answers · asked by lil Konvict 4 in Science & Mathematics ➔ Mathematics
put the numbers as a fraction and work out what the factors of the numbers are
1156/289 = (2*2*17*17) / (17*17)
Then you see that you can cancel the 17's and what you are really after is SQRT(2*2) which is 2
0.000049 = 49/1000000
and this in turn is (7*7) / (2*2*2*2*2*2*5*5*5*5*5*5)
and so you are after SQRT(7*7) / SQRT(10*10*10*10*10*10) which is 7/1000 or 0.007
2007-04-16 20:23:45 · answer #1 · answered by Orinoco 7 · 1⤊ 0⤋
Question 1
√(1156/289) = √4 = 2
Question 2
√0.000049 = 0.007 where:-
√0.000049 = √49 x √10^(-6)
= 7 x 10^(-3)
= 0.007
2007-04-16 22:39:11 · answer #2 · answered by Como 7 · 0⤊ 0⤋
In monomial expressions, when taking the square root of a variable, divide the exponent by 2. Since a^2bc^11 is a monomial, √(a^2bc^11) = [a^(2/2)][b^(1/2)][c^(11/2)] = a√b√c^(11)
2016-05-17 06:51:31 · answer #3 · answered by ? 3 · 0⤊ 0⤋
a. 1156/289=4 ..so ...sqrt(4)=2
b. 49/1000000=(7/1000)^2 so...sqrt(0.000049)=0.007
2007-04-16 22:57:44 · answer #4 · answered by Anonymous · 0⤊ 0⤋