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11. x = 1/2 , y = 1/3
12. if y =1
line is parallel to x - axis
13. 2x^2-5x+3 = 0
2x^2 -2x-3x+3 =0
2x(x-1)-3(x-1) = 0
(2x-3)(x-1) =0
x= 3/2 , 1
14 x = -b+-√(b^2-4ac)/2a
given b^2-4ac = 0
x = -b/2a , -b/2a
15. x + 1/x = 10/3 = 3+ 1/3
by comparing
x = 3
17. Term = a+(n-1) d = -22
5+(n-1) (-3) = -22
5-3n+3 = -22
8-3n = -22
3n =30
n=10
18 cot π/3 = 1/√3
cot π/4 =1
cot π/3 = √3
There is a constant ratio of √3 in each consecutive term so it in G.P.
19. Distance = √{(-4-5)^2+(-1-2)^2} = √81+9 = √90
20. By section formula
x = m1x2+m2x1/m1+m2
y = m1y2+m2y1/m1+m2
Q= { (7/2*2+4*1)/(2+1) , (9/2*2+2*1)/(2+1)}
Q= (11/3 , 11/3)
12. if y =1
line is parallel to x - axis
13. 2x^2-5x+3 = 0
2x^2 -2x-3x+3 =0
2x(x-1)-3(x-1) = 0
(2x-3)(x-1) =0
x= 3/2 , 1
14 x = -b+-√(b^2-4ac)/2a
given b^2-4ac = 0
x = -b/2a , -b/2a
15. x + 1/x = 10/3 = 3+ 1/3
by comparing
x = 3
17. Term = a+(n-1) d = -22
5+(n-1) (-3) = -22
5-3n+3 = -22
8-3n = -22
3n =30
n=10
18 cot π/3 = 1/√3
cot π/4 =1
cot π/3 = √3
There is a constant ratio of √3 in each consecutive term so it in G.P.
19. Distance = √{(-4-5)^2+(-1-2)^2} = √81+9 = √90
20. By section formula
x = m1x2+m2x1/m1+m2
y = m1y2+m2y1/m1+m2
Q= { (7/2*2+4*1)/(2+1) , (9/2*2+2*1)/(2+1)}
Q= (11/3 , 11/3)
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