Class 12th
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New answer posted
4 months agoContributor-Level 10
adj A| = |A|² = 9
=> |A| = ±3 => λ = |λ| = 3
=> |B| = |adj A|² = 81
=> | (B? ¹)? | = |B? ¹| = |B|? ¹ = 1/|B| = 1/81 = µ
New answer posted
4 months agoContributor-Level 10
Mid point P = (3,1,1)
Normal of plane is along the line AB.
D.R.'s of normal = 4-2, -2-4, 3-1 (-1) = 2, -6, 4
= 1, -3, 2
Plane -> 1 (x-3) - 3 (y-1) + 2 (z-1) = 0
=> x - 3y + 2z - 2 = 0
New answer posted
4 months agoContributor-Level 10
m? = 6/-12 = -1/2
∴ Equation of AD is y-7 = 2 (x+1)
y = 2x+9
m? = 12/-6 = -2
∴ Equation of BE is y-1 = 1/2 (x+7)y
y = x/2 + 9/2
by (1) and (2)
2x+9 = (x+9)/2
=> 4x+18 = x+9
=> 3x = 9 => x=-3
∴ y=3
New answer posted
4 months agoContributor-Level 10
S = 6a² => dS/dt = 12a * da/dt = 3.6
=> 12 (10) da/dt = 3.6
=> da/dt = 0.03
V = a³ => dV/dt = 3a² * da/dt
= 3 (10)² * (3/100) = 9
New answer posted
4 months agoContributor-Level 10
Total = 9 (10? )
Fav. Way =? C? (2? -2) +? C? (2? -1) = 36 (30) + 9 (15) = 1080 + 135
Probability = (36x30+9x15)/ (9x10? ) = (4x30+15)/10? = 135/10?
New answer posted
4 months agoContributor-Level 10
x³dy + xydx = 2ydx + x²dy
=> (x³-x²)dy = (2-x)ydx
=> dy/y = (2-x)/ (x² (x-1) dx
=> ∫ (dy/y) = ∫ (2-x)/ (x² (x-1)dx
Let (2-x)/ (x² (x-1) = A/x + B/x² + C/ (x-1)
=> 2-x = A (x-1) + B (x-1) + Cx²
=> C=1, B=-2 and A=-1
=> ∫ (dy/y) = ∫ (-1/x - 2/x² + 1/ (x-1)dx
=> lny = -lnx + 2/x + ln|x-1| + C
∴ y (2) = e
=> 1 = -ln2 + 1 + 0 + C
=> C = ln2
=> lny = -lnx + 2/x + ln|x-1| + ln2
at x = 4
=> lny (4) = -ln4 + 1/2 + ln3 + ln2
=> lny (4) = ln (3/4) + 1/2 = ln (3/2)e^ (1/2)
=> y (4) = (3/2)e^ (1/2)
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