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New answer posted
2 months agoContributor-Level 10
Let five terms in G.P. be a/r², a/r, a, ar, ar²
Then, a (r? ² + r? ¹ + 1 + r + r²) / (1/a) (r? ² + r? ¹ + 1 + r? ¹ + r? ²) = 49
⇒ a² = 49 ⇒ a = ±7
Also, a/r² + a = 35
Therefore, a = -7 is not possible
Now, fifth term = ar² = a (7/28) = p ⇒ 4p = 7
New answer posted
2 months agoContributor-Level 10
T? = cot? ¹ (2²? ¹ + 1/2? ) = cot? ¹ (1 + 2²? ¹)/2? ) = tan? ¹ (2? / (1 + 2²? ¹)
T? = tan? ¹ (2? ¹ - 2? ) / (1 + 2? ¹ . 2? ) = tan? ¹ (2? ¹) – tan? ¹ (2? )
S∞ = π/2 - tan? ¹ (2) = cot? ¹ (2) = tan? ¹ (1/2)
New answer posted
2 months agoContributor-Level 10
3x² + 3x²y - 3xy² + dy³ = 0
3x² (x + y) – 3y² (x - dy/3) = 0
x - dy/3 = x + y for getting two perpendicular straight lines
d = -3
New answer posted
2 months agoContributor-Level 10
cos 20° cos 40° cos 80°
= (2 sin 20° cos 20° cos 40° cos 80°) / (2 sin 20°)
= (2 sin 40° cos 40° cos 80°) / (2² sin 20°)
= (2 sin 80° cos 80°) / (2³ sin 20°)
= 1/8
New answer posted
2 months agoContributor-Level 10
lim (x→1) [f (1)g (x)-f (1)-g (1)f (x)+g (1)] / [f (1)g (x)-f (x)g (1)], form: 0/0
lim (x→1) [f (1)g' (x)-g (1)f' (x)] / [f (1)g' (x)-f' (x)g (1)] = 1
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