Class 11th
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New answer posted
6 months agoContributor-Level 10
The equation of the tangent to the ellipse x²/27 + y² = 1 at the point (3√3 cosθ, sinθ) is:
x (3√3 cosθ)/27 + y (sinθ)/1 = 1 ⇒ x/ (3√3) cosθ + y sinθ = 1.
To find the intercepts on the axes:
x-intercept (set y=0): x = 3√3 / cosθ = 3√3 secθ.
y-intercept (set x=0): y = 1 / sinθ = cosecθ.
The sum of the intercepts is z (θ) = 3√3 secθ + cosecθ.
To find the minimum value of z, we differentiate with respect to θ and set it to zero:
dz/dθ = 3√3 secθ tanθ - cosecθ cotθ = 0.
3√3 (sinθ/cos²θ) = cosθ/sin²θ.
3√3 sin³θ = cos³θ ⇒ tan³θ = 1/ (3√3).
⇒ tanθ = 1/√3.
Since θ ∈ (0, π/2), the solution
New answer posted
6 months agoContributor-Level 10
The problem provides non-standard formulas for sums S? and S? :
S? = n [a + (2n-1)d]
S? = 2n [a + (4n-1)d]
We are given S? - S? = 1000.
S? - S? = 2n [a + (4n-1)d] - n [a + (2n-1)d]
= (2na + 8n²d - 2nd) - (na + 2n²d - nd)
= na + 6n²d - and = n [a + (6n-1)d].
So, n [a + (6n-1)d] = 1000.
We need to find S? n. Assuming the pattern S? n = kn [a + (2kn-1)d], then S? n would be 3n [a + (6n-1)d].
S? n = 3 * (n [a + (6n-1)d]) = 3 * 1000 = 3000.
New answer posted
6 months agoContributor-Level 10
Given the equation 15 sin? + 10 cos? = 6.
Divide by cos? : 15 tan? + 10 = 6 sec?
Using sec²? = 1 + tan²? , we get sec? = (1 + tan²? )² = 1 + 2tan²? + tan?
15 tan? + 10 = 6 (1 + 2tan²? + tan? ).
15 tan? + 10 = 6 + 12tan²? + 6tan?
9 tan? - 12 tan²? + 4 = 0.
This is a quadratic in tan²? : (3 tan²? - 2)² = 0.
? 3 tan²? = 2? tan²? = 2/3.
From this, we find sin²? and cos²? If tan²? = 2/3, then sin²? = 2/5 and cos²? = 3/5.
Also, sec²? = 1 + tan²? = 5/3 and cosec²? = 1 + cot²? = 1 + 3/2 = 5/2.
The expression to evaluate is 27 sec? + 8 cosec? = 27 (sec²? )³ + 8 (cosec²? )³.
= 27 (5/3)³ + 8 (5/2)³ = 27 (125/27) + 8 (125/8) =
New answer posted
6 months agoContributor-Level 9
Oxidation state of N
| NO | +2 |
|-|-|
| NO? | +4 |
| N? O | +1 |
| NO? | +5 |
So, order of oxidation state is
NO? > NO? > NO > N? O
New answer posted
6 months agoContributor-Level 10
The data consists of n values of a and n values of -a.
Mean x? = (n*a + n* (-a) / 2n = 0 / 2n = 0.
Variance σ² = (Σx? ²)/N - x? ² = (n*a² + n* (-a)²) / 2n - 0² = 2na² / 2n = a².
If a value b is added to all observations, the new mean is x? ' = x? + b = 0 + b = b.
We are given the new mean is 5, so b=5.
Adding a constant does not change the variance. The new variance is still a².
We are given the new standard deviation is 20, so the new variance is 20² = 400.
Thus, a² = 400.
The required value is a² + b² = 400 + 5² = 400 + 25 = 425.
New answer posted
6 months ago
Contributor-Level 9
Hybridisation of carbon a, b, and c respectively are sp³, sp² and sp².
New answer posted
6 months agoContributor-Level 10
The equation is for a hyperbola: x²/4 - y²/2 = 1.
The eccentricity e is given by e = √ (1 + b²/a²) = √ (1 + 2/4) = √6/2.
The focus F is at (ae, 0), which is (2 * √6/2, 0) = (√6, 0).
The equation of the tangent at a point P (x? , y? ) is xx? /a² - yy? /b² = 1.
The equation of the tangent at P is given as x - (√6 y)/2 = 1.
This tangent cuts the x-axis (y=0) at x=1, so Q is (1, 0).
The latus rectum is the line x = ae = √6.
To find the point R where the tangent intersects the latus rectum, we substitute x=√6 into the tangent equation:
√6 - (√6 y)/2 = 1 ⇒ √6 - 1 = (√6 y)/2 ⇒ y = 2 (√6 - 1)/√6.
The vertices o
New answer posted
6 months agoContributor-Level 10
Given a right-angled triangle with one angle π/3 (or 60°), and the side adjacent to it is h/2.
Using the tangent function: tan (π/3) = opposite / adjacent.
Let the adjacent side be 2. Then tan (π/3) = h/2.
We know tan (π/3) = √3.
So, h/2 = √3, which implies h = 2√3.
New answer posted
6 months agoContributor-Level 10
Given a triangle with inradius r and circumradius R.
The inradius r is calculated as r = |0 + 0 - 3| / √2 = 3 / √2.
From the geometry of the triangle, we have the relation r/R = sin (30°) = 1/2.
This gives R = 2r.
The question asks for the sum R + r, which is 2r + r = 3r.
Substituting the value of r, we get 3 * (3/√2) = 9/√2.
∴ R + r = 9/√2.
New answer posted
6 months agoContributor-Level 10
(1/2)kx² = (1/2)mv² ⇒ x = v√ (m/k) = 10 * √ (4/100) = 10 * (2/10) = 2m
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