Given that
log (x+y)/5 =( 1/2) {log x+logy}
We know that
log a+ log b = log ab
⇛log (x+y)/5 =( 1/2) log(xy)
We know that log a^m = m log a
⇛log (x+y)/5 = log (xy)^1/2
⇛log (x+y)/5 = log√(xy)
⇛(x+y)/5 = √(xy)
On squaring both sides then
⇛{ (x+y)/5}^2 = {√(xy)}^2
⇛(x+y)^2/5^2 = xy
⇛(x^2+y^2+2xy)/25 = xy
⇛x^2+y^2+2xy = 25xy
⇛x^2+y^2 = 25xy-2xy
⇛x^2+y^2 = 23xy
⇛( x^2+y^2)/xy = 23
⇛(x^2/xy) +(y^2/xy) = 23
⇛{(x×x)/xy} +{(y×y)/xy} = 23
⇛(x/y)+(y/x) = 23
Therefore, (x/y)+(y/x) = 23
Hence, the value of (x/y)+(y/x) is 23.
-1 is less than 0, so you use the first equation:
3(-1) +2 = -3+2 = -1
f(-1) = -1
For 0 use the 2nd equation:
3(0) + 4 = 0+4 = 4
f(0) = 4
For 2 use the 2nd equation:
3(2) + 4 = 6+4 = 10
f(2) = 10
In this case we use intersecting secant theorem
According to this theorem, GF*GE = GH*GA
So,we plug in the values
(x-3)*(x-3+9) =(x-2)*(x-2+6)
(x-3)*(x+6) =(x-2)*(x+4)
x^2 +6x -3x -18 =x^2 +4x -2x -8
x^2 +3x -18 =x^2 +2x -8
x^2 +3x -18 -x^2 -2x +8 =0
x -10 =0
x=10
So,FG =x-3 = 10-3 =7
Answer:
Students 1 and 2
Step-by-step explanation:
