Answer:
The ratio in which the line x-y-2=0 divides the line joining A(3,-1) and B(8,9) is in the ratio 2:3
Step-by-step explanation:
Let (x, y) be the coordinates of point of intersection.
Hence x=(a*8+1*3)(a+1) = (8a+3)/(a+1)
and
y = {a*9+1*(-1)}/(a+1)=(9a-1)/(a+1)
Since this point lies on the line x-y-2=0
Hence (8a+3)/(a+1)-(9a-1)/(a+1)-2=0
i.e. 8a+3–9a+1–2(a+1)=0
Or 8a+3–9a+1–2a-2=0
i.e.-3a+2=0
Hence a=2/3
hence the ratio in which the line x-y-2=0 divides the line joining A(3,-1) and B(8,9) in the ratio 2/3:1
i.e. 2:3
Answer:ok
Step-by-step explanation:
Answer:
y = 3sin2t/2 - 3cos2t/4t + C/t
Step-by-step explanation:
The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt
Comparing the standard form with the given differential equation.
p(t) = 1/t and q(t) = 3cos(2t)
I = e^∫1/tdt
I = e^ln(t)
I = t
The general solution for first a first order DE is expressed as;
y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.
yt = ∫t(3cos2t)dt
yt = 3∫t(cos2t)dt ...... 1
Integrating ∫t(cos2t)dt using integration by part.
Let u = t, dv = cos2tdt
du/dt = 1; du = dt
v = ∫(cos2t)dt
v = sin2t/2
∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt
= tsin2t/2 - cos2t/4 ..... 2
Substituting equation 2 into 1
yt = 3(tsin2t/2 - cos2t/4) + C
Divide through by t
y = 3sin2t/2 - 3cos2t/4t + C/t
Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t
Answer:
something perpendicular probably
Since 68% of the sample ranges from the plus minus first standard deviation from the mean, then interval wherein it will occur will be between 53.5% to 60.5%. I got this answer through adding and subtracting one standard deviation (3.5) from the mean (which is 57).
