Class 11th
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New question posted
2 months agoNew answer posted
2 months agoContributor-Level 10
u = (2z+i)/ (z-ki)
= (2x² + (2y+1) (y-k)/ (x²+ (y-k)²) + I (x (2y+1) - 2x (y-k)/ (x²+ (y-k)²)
Since Re (u) + Im (u) = 1
⇒ 2x² + (2y+1) (y-k) + x (2y+1) - 2x (y-k) = x² + (y-k)²
P (0, y? )
Q (0, y? )
⇒ y² + y - k - k² = 0 {y? + y? = -1, y? = -k-k²}
∴ PQ = 5
⇒ |y? - y? | = 5 ⇒ k² + k - 6 = 0
⇒ k = -3, 2
So, k = 2 (k > 0)
New answer posted
2 months agoContributor-Level 10
Let TV (r) denotes truth value of a statement r.
Now, if TV (p) = TV (q) = T
⇒ TV (S? ) = F
Also, if TV (p) = T and TV (q) = F
⇒ TV (S? ) = T
New answer posted
2 months agoContributor-Level 10
1 + (1 - 2²⋅1) + (1 - 4²⋅3) + . + (1 - 20²⋅19)
= α - 220β
= 11 - (2²⋅1 + 4²⋅3 + . + 20²⋅19)
= 11 - 2² ⋅ Σ? (r=1) r² (2r-1) = 11 - 4 (110²/2) - 35 x 11)
= 11 - 220 (103)
⇒ α = 11, β = 103
New answer posted
2 months agoContributor-Level 10
x? = 10
⇒ x? = (63 + a + b)/8 = 10
⇒ a + b = 17
Since, variance is independent of origin.
So, we subtract 10 from each observation.
So, σ² = 13.5 = (79 + (a-10)² + (b-10)²)/8
⇒ a² + b² - 20 (a+b) = -171
⇒ a² + b² = 169
From (1) and (2) ; a = 12 and b = 5
New answer posted
2 months agoContributor-Level 10
x²/a² + y²/b² = 1 (a > b); 2b²/a = 10 ⇒ b² = 5a
Now, φ (t) = -5/12 + t - t² = 8/12 - (t - 1/2)²
φ (t)max = 8/12 = 2/3 = e ⇒ e² = 1 - b²/a² = 4/9
⇒ a² = 81 (From (i) and (ii)
So, a² + b² = 81 + 45 = 126
New answer posted
2 months agoContributor-Level 10
x² - 3x + p = 0
α, β, γ, δ in G.P.
α + αr = 3
x² - 6x + q = 0
ar² + ar³ = 6
(2) ÷ (1) ⇒ r² = 2
So, 2q+p/2q-p = (2r? +r)/ (2r? -r) = (2r? +1)/ (2r? -1) = 9/7
New answer posted
2 months agoContributor-Level 10
Angular momentum conservation:
⇒ I? ω? + I? ω? = (I? + I? )ωf
⇒ (MR²/2)ω? = (MR²/2 + MR²/8)ωf
⇒ ωf = 4/5 ω?
⇒ KEfinal = ½ (I? + I? )ωf² = (MR²ω? ²)/5
⇒ KEinitial = ½I? ω? ² = (MR²ω? ²)/4
⇒ % loss ⇒ 20%
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