mohanrao-d
mohanrao-d
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cot^2(x) + cos(x) - csc(x) cot( x) = { cos^2(x) / sin^2(x) } + cos x - { cos x / sin^2(x) } = { cos^2(x) + cos x sin^2(x) - cos x ) / sin^2(x) = cos x [ cos x + sin^2(x) - 1 ] / sin^2(x) = cos x [ cos x - cos^2(x) ] / sin^2(x) = …
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The first given condition is sum of two numbers is 8 so x + y = 8, where x and y are positive numbers y = 8 - x second condition is x^3 + y^2 is minimum replacing y with 8-x x^3 + (8-x)^2 is minimum x^3 + x^2 - 16x + 64 find the…
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The first given condition is sum of two numbers is 8 so x + y = 8, where x and y are positive numbers y = 8 - x second condition is x^3 + y^2 is minimum replacing y with 8-x x^3 + (8-x)^2 is minimum x^3 + x^2 - 16x + 64 find the…
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let the assistant can decorate in x minutes so in 1 minute the assistant can decorate -------1/ x part in 1 minute baker can decorate --------------------4/x part so in 1 minute both together can decorate ----(1/x) + (4/x) = 5/x part Bot…
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let the assistant can decorate in x minutes so in 1 minute the assistant can decorate -------1/ x part in 1 minute baker can decorate --------------------4/x part so in 1 minute both together can decorate ----(1/x) + (4/x) = 5/x part Bot…
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let arcsin(2x) = A sin A = 2x cos A = SQRT(1 - sin^2(A)) = SQRT(1 - 4x^2) => cos [arcsin(2x)] = cos A = SQRT(1 - 4x^2)
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let the time taken for them to be 1380 miles apart = t hours The distance traveled by Rush = time* rate = 600t miles The distance traveled by Slow = 320t Thus 600t + 320t = 1380 920t = 1380 t = 1380/920 = 1.5 hours
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1) first two digit number ending with 4 = 14 last two digit number ending with 4 = 94 so the series is 14,24,34.........94 a1 = 14, d = 10, n = 9 S(9) = n/2[a1+an] = 9/2[14+94] = 486 2) first three digit number divisible by …
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20u1 + 190d = 540 divide by 10 2u1 + 19d = 54-----------------------eqn(1) 1260 = 30u1 + 435d divide by 15 2u1 + 29d = 84----------------------eqn(2) subtract (1) from (2) 10d = 30 d = 3 substitute d value in eqn (1) 2u…
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20u1 + 190d = 540 divide by 10 2u1 + 19d = 54-----------------------eqn(1) 1260 = 30u1 + 435d divide by 15 2u1 + 29d = 84----------------------eqn(2) subtract (1) from (2) 10d = 30 d = 3 substitute d value in eqn (1) 2u…
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S_n = a(r^n - 1)/(r - 1) S_10 = 5(2^10 - 1) = 5(1024 - 1) = 5115
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3/4(5x - 1/2) - 5/3 < 1/3 add (5/3) to both sides 3/4(5x - 1/2) < 2 multiply both sides with 4 3(5x - 1/2) < 8 15x - 3/2 < 8 add 3/2 to both sides 15x < 19/2 divide both sides by 15 x < 19/30
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y = 10(x^5) e^(xsinx) let u = x^5 : du = 5x^4 v = e^(x sinx) : dv = e^(x sinx)*(x cosx + sinx ) d(uv) = udv + vdu d[10(x^5) e^(xsinx)] = 10 x^5*(x cosx + sinx)*e^(x sinx) + 50 x^4* e^(x sinx)
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y = 10(x^5) e^(xsinx) let u = x^5 : du = 5x^4 v = e^(x sinx) : dv = e^(x sinx)*(x cosx + sinx ) d(uv) = udv + vdu d[10(x^5) e^(xsinx)] = 10 x^5*(x cosx + sinx)*e^(x sinx) + 50 x^4* e^(x sinx)
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3/4(5x - 1/2) - 5/3 < 1/3 add (5/3) to both sides 3/4(5x - 1/2) < 2 multiply both sides with 4 3(5x - 1/2) < 8 15x - 3/2 < 8 add 3/2 to both sides 15x < 19/2 divide both sides by 15 x < 19/30
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3/4(5x - 1/2) - 5/3 < 1/3 add (5/3) to both sides 3/4(5x - 1/2) < 2 multiply both sides with 4 3(5x - 1/2) < 8 15x - 3/2 < 8 add 3/2 to both sides 15x < 19/2 divide both sides by 15 x < 19/30
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S_n = a(r^n - 1)/(r - 1) S_10 = 5(2^10 - 1) = 5(1024 - 1) = 5115
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20u1 + 190d = 540 divide by 10 2u1 + 19d = 54-----------------------eqn(1) 1260 = 30u1 + 435d divide by 15 2u1 + 29d = 84----------------------eqn(2) subtract (1) from (2) 10d = 30 d = 3 substitute d value in eqn (1) 2u…
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20u1 + 190d = 540 divide by 10 2u1 + 19d = 54-----------------------eqn(1) 1260 = 30u1 + 435d divide by 15 2u1 + 29d = 84----------------------eqn(2) subtract (1) from (2) 10d = 30 d = 3 substitute d value in eqn (1) 2u…
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20u1 + 190d = 540 divide by 10 2u1 + 19d = 54-----------------------eqn(1) 1260 = 30u1 + 435d divide by 15 2u1 + 29d = 84----------------------eqn(2) subtract (1) from (2) 10d = 30 d = 3 substitute d value in eqn (1) 2u…
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20u1 + 190d = 540 divide by 10 2u1 + 19d = 54-----------------------eqn(1) 1260 = 30u1 + 435d divide by 15 2u1 + 29d = 84----------------------eqn(2) subtract (1) from (2) 10d = 30 d = 3 substitute d value in eqn (1) 2u…
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1) first two digit number ending with 4 = 14 last two digit number ending with 4 = 94 so the series is 14,24,34.........94 a1 = 14, d = 10, n = 9 S(9) = n/2[a1+an] = 9/2[14+94] = 486 2) first three digit number divisible by …
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let the time taken for them to be 1380 miles apart = t hours The distance traveled by Rush = time* rate = 600t miles The distance traveled by Slow = 320t Thus 600t + 320t = 1380 920t = 1380 t = 1380/920 = 1.5 hours
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let the time taken for them to be 1380 miles apart = t hours The distance traveled by Rush = time* rate = 600t miles The distance traveled by Slow = 320t Thus 600t + 320t = 1380 920t = 1380 t = 1380/920 = 1.5 hours
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let the assistant can decorate in x minutes so in 1 minute the assistant can decorate -------1/ x part in 1 minute baker can decorate --------------------4/x part so in 1 minute both together can decorate ----(1/x) + (4/x) = 5/x part Bot…
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The first given condition is sum of two numbers is 8 so x + y = 8, where x and y are positive numbers y = 8 - x second condition is x^3 + y^2 is minimum replacing y with 8-x x^3 + (8-x)^2 is minimum x^3 + x^2 - 16x + 64 find the…