_/128 - _/18
how do u solve this????????
plz hlp!!!!
2006-11-30 15:39:32 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
_/128 = SQRT(128) = SQRT(64 * 2)
= 8 * SQRT(2)
Similarly _/18 = 3 * SQRT(2)
=> _/128 - _/18 = 8 * SQRT(2) - 3 * SQRT(2)
= 5 * SQRT(2)
Answer : 5_/2
2006-11-30 15:46:36 · answer #1 · answered by Pavan 2 · 0⤊ 0⤋
OK. Do it like this;
You know that 2x64=128 right. So, put them both under a sq root sign. The 64 has a sq root and it is 8, so put an 8 in front of the sq root symbol and leave the 2 under it because it does not have a perfect square.
Now for the sq root of 18. Same thing--break it down into a 9 and a 2. The 9 has a perfect square and it is a 3. So put a three in front of this sq root symbol and take out the 9 and leave the 3 under there. You end up with 8_/2 - 3_/2=5_/2
2006-11-30 23:49:30 · answer #2 · answered by Tony T 4 · 0⤊ 0⤋
Are _/ square root symbols?
â128-â18
8â2-3V2
5â2
2006-11-30 23:47:18 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
since they have'nt exact square roots, we factor them to the numbers that has exact roots.
2*64=128 and 2*9=18
_/64=8 and _/9=3
so;
= _/2*64 - _/2*9
= 8_/2 - 3_/2
= 5_/2
2006-12-01 05:34:56 · answer #4 · answered by sheryl 1 · 0⤊ 0⤋
if the _/ is a sq root sign. then
_/128= 8_/2
_/18=3_/2
8_/2 - 3_/2 = 5_/2
i hope this helps
2006-12-01 00:16:58 · answer #5 · answered by LOL-sweety_pie-☆ 1 · 0⤊ 0⤋