2006-11-21 11:33:44 · 12 answers · asked by Yazzy 1 in Science & Mathematics ➔ Mathematics
x+y=7
xy=25
1/x+1/y=?
if you add by making both denominators equal (gcf is xy) you get
(y+x)/(xy)=7/25
2006-11-21 11:43:45 · answer #1 · answered by rawfulcopter adfl;kasdjfl;kasdjf 3 · 0⤊ 1⤋
1/2 + 1/5 = 7/ 10 =.7
2006-11-21 12:01:18 · answer #2 · answered by Anonymous · 1⤊ 1⤋
x+y=7 ; x=7-y ;
xy=25; y=25/(7-y)
The sum of reciprocals 1/x+1/y=1/(7-y)+1/[25/(7-y)];
1/(7-y) + (7-y)/25
2006-11-21 12:10:07 · answer #3 · answered by Anonymous · 1⤊ 1⤋
x+y = 7
x*y = 25
x = 7-y
(7-y)y = 25
7y - y^2 = 25
y^2 -7y +25 = 0
y = ((7 +or- sqrt(49-4*25*1))/2
y = (7 +or- sqrt(-51))/2
y = (7 + i*sqrt51)/2 , (7 - i*sqrt51)/2
y = 3.5 + i*(sqrt51)/2, 3.5 - i*(sqrt51)/2
x = 3.5 - i*(sqrt51)/2, 3.5 + i*(sqrt51)/2
1/x + 1/y = sum of reciprocals
2/(7 + i*sqrt51) + 2/(7 - i*sqrt51) = sum
(2(7 - i*sqrt51) +2(7 + i*sqrt51))/(7 + i*sqrt51)(7-i*sqrt51) = sum
28/(49 +51) = sum
28/100 = 7/25 = .28 = sum
2006-11-21 11:58:28 · answer #4 · answered by Steve A 7 · 1⤊ 1⤋
the only thing i can think of is 5 ^ 2. so .2 + .5 = .7
2006-11-21 11:39:30 · answer #5 · answered by Mike 2 · 1⤊ 1⤋
1, reciprocals always =1
2006-11-21 11:36:49 · answer #6 · answered by P!ATD GIRL! Luvn Brendon Urie! 3 · 0⤊ 2⤋
I tried a bunch of stuff, substitution, factoring, quadratics, and got that one of the #'s is an imaginary number, this is too hard, I quit.
2006-11-21 11:45:59 · answer #7 · answered by quickster94 3 · 0⤊ 1⤋
x+y=7
xy = 25
(1/x) + (1/y) = (x+y)/(xy) = 7/25
(
google built-in calculator tells us
((7 + (i * sqrt(51))) / 2) +
((7 - (i * sqrt(51))) / 2) = 7
((7 + (i * sqrt(51))) / 2) *
((7 - (i * sqrt(51))) / 2) = 25
(1 / ((7 + (i * sqrt(51))) / 2)) +
(1 / ((7 - (i * sqrt(51))) / 2)) = 0.28
)
2006-11-21 12:20:50 · answer #8 · answered by paladin 1 · 1⤊ 1⤋
(1/x) +(1/y) = (x+y)/xy = 7/25
2006-11-21 11:41:00 · answer #9 · answered by Anonymous · 3⤊ 1⤋
Boooooo math. I hate math. So I don't care about the correct answer. lol
2006-11-21 11:43:33 · answer #10 · answered by Duct Girlie 2 · 0⤊ 2⤋