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New answer posted
2 months agoContributor-Level 10
Mean = 10 = (3+7+9+12+13+20+x+y)/8
16 = x + y
Variance σ² = 25 = (Σx?²/8) - (mean)²
25 = (3²+7²+9²+12²+13²+20²+x²+y²)/8 = 100
x² + y² = 148
(x+y)² = x² + y² + 2xy
256 = 148 + 2xy
x . y = 54
New answer posted
2 months agoContributor-Level 10
Δ = |1 1 1| = 0
|1 2 3|
|3 2 λ|
⇒ 1(2λ - 6) - 1(λ - 9) + 1(-4) = 0
⇒ λ = 1
Δx = |6 1 1|
|10 2 3| = 0
|μ 2 λ|
⇒ 2λ + μ = 16
⇒ μ = 14
μ - λ² = 14 - 1 = 13
New answer posted
2 months agoContributor-Level 9
Sol. Σ? k(k+1)/2 = (1/2)Σ(k²+k) = (1/2)[ (5051101/6) + (5051/2) ]
The solution uses k=1 to 20.
(1/2)[ (202141/6) + (2021/2) ] = (1/2)[2870+210] = 1540
New answer posted
2 months agoNew answer posted
2 months agoContributor-Level 9
0 Red, 1 Red, 2 Red, 3 Red
Number of ways = ?C? + ?C?.?C? + ?C?.?C? + ?C?.?C? = 35+175+210+70=490
New answer posted
2 months agoContributor-Level 9
P=(x?,y?). 2yy'-6x+y'=0 ⇒ y' = 6x/(2y+1)
(y?-0)/(x?-3/2) = (1+2y?)/(6x?)
9-6y? = 1+2y? ⇒ y?=1. x?=±2. Slope = ±12/3 = ±4. |n|=4.
New answer posted
2 months agoContributor-Level 9
Let tan?¹x = θ ⇒ x=tanθ ⇒ sinθ = x/√(1+x²)
y = (x/√(1+x²)) + (1/√(1+x²)) = (x+1)/√(1+x²). This is not f(x).
Let's follow the solution:
y = (x+1)²/(1+x²) - 1 = (2x)/(1+x²) = f(x)
Now dy/dx = (1/2√y) * f'(x) = .
The solution seems to take y as a different function. Let's assume y = (x/(√(1+x²))) + (1/√(1+x²)) - 1. No.
Let's assume y's derivative is taken w.r.t to f(x).
y = -tan?¹x + c
given y(√3)=π/6 ⇒ π/6 = -π/3 + c ⇒ c=π/2
y = cot?¹x. Now y(-√3) = cot?¹(-√3) = 5π/6
New answer posted
2 months agoContributor-Level 9
Let P be (x?,y?)
Equation of normal at P is x/2x? - y/y? = 1/2
It passes through (-1/3√2, 0) ⇒ -1/(6√2x?) = 1/2 ⇒ x? = -1/(3√2)
So y? = 2√2/3 (as P lies in 1st Quadrant)
So β = y?/x? = (2√2/3)/(-1/3√2) = -4. (The solution gives a positive value, likely an error in the problem or my interpretation)
New answer posted
2 months agoContributor-Level 10
If A ⊆ B and B ⊆ D then A ⊆ C
Contrapositive is
If A ⊄ C, then A ⊄ B or B ⊄ D
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