Question

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Question. 10 point
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Q.) If the radius of the base of a cylinder is increased by 20% and its height is increased by 10%, then its volume will be increased by how many percent?

Please someone thank my all answer.(╯︵╰,)​

Answer 1

The percentage change in the volume is 6.4%.

Step-by-step explanation:

Given : If the radius of a cylinder is increased by 20% and its height is decreased by 10%

Answer 2

Hey there, here is your answer of this Question ;)

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$\underline{\fbox{ {☆ \red{Required \: solution}☆} }}$

$ɪ = \sf\int \frac{x {}^{2} + 3x + 2}{(x {}^{2} + 1) {}^{2}(x + 1) } dx \\ \\ \sf ɪ = \int \frac{(x {}^{2} + 2x) + (1x + 2)}{(x {}^{2} + 1) {}^{2}(x + 1) {}^{2} } \\ \\ \sf \: ɪ = \int \frac{(x + 2) \red{(x + 1)}}{(x {}^{2} + 1) {}^{2} \red{(x + 1)} } \\ \\ ɪ = \sf \int \frac{x + 2}{(x {}^{2} + 1) {}^{2} } dx \\ \\ ɪ = \sf \int \frac{x}{(x {}^{2} + 1) {}^{2} } dx + \int \frac{2}{(x {}^{2} + 1) {}^{2} } dx \\ \\ \boxed{\boxed{ \green{ \sf ɪ = ɪ_{1} + ɪ_{2}}}} - - - (1) \\ \\ ɪ_{1} = \int \sf \frac{x}{(x {}^{2} + 1) {}^{2} } dx \\ \\ \boxed{\red{ \sf \: put \: x {}^{2} = t} } \\ \red{ \: 2x \: dx = dt }\\ \red{x \: d_{x} = \frac{1}{2}dt } \\ \\ ɪ _{1} = \sf \frac{1}{2} \int \frac{dt}{(t + 1) {}^{2} } \\ \\ ɪ _{1} = \sf\frac{1}{2} \frac{(t + 1) {}^{ - 1} }{ - 1} + 0 \\ \\ ɪ _{1} = \sf \frac{ - 1}{2(x { }^{2} + 1)} = 0$

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$ɪ _{2} = \sf \int \frac{2dx}{(x {}^{2} + 1) {}^{2} } \\ \red{put \: x = \tan \theta} \\ \red{ \sf \: dx = \sec {}^{2} \theta \: d \theta} \\ \\ ɪ _{2} = 2 \sf \int \frac{ \sec {}^{2} \theta \: d \theta }{(1 + \tan {}^{2} \theta) {}^{2} } \\ \\ ɪ _{2} = \sf2 \int \frac{ \sec {}^{2} \theta \: d \theta }{( \sec {}^{2} \theta) {}^{2} } \\ \\ ɪ _{2} = 2\int \sf\frac{ \sec {}^{2} d \theta }{ \sec {}^{4} \theta} \\ \\ ɪ _{2} = \int \sf \frac{1 d \theta}{ \sec {}^{2} \theta} \\ \\ ɪ _{2} = \sf 2\int \cos {}^{2} \theta d\theta \\ \\ \boxed{ \int2 \cos {}^{2} \theta \: d \theta = 2 \cos {}^{2} \theta} \\ \\ ɪ _{2} = \int \sf(1 + \cos {}^{2} \theta) d \theta \\ \\ ɪ _{2} = \frac{ \theta + \sin2 \theta}{2} + c _{2} \\ \\ \boxed{ \red{x = tan \theta}} \\ \\ \sin {}^{2} \theta = \frac{2 \tan \theta}{1 + \tan {}^{2} \theta } \\ \\ \sf \sin {}^{2} \theta = \frac{2x}{1 + x {}^{2} } \\ \\ ɪ _{2} = \sf \tan { }^{ - 1} x + \frac{x}{1 + x {}^{2} } + c_{2}$

Put in Equation 1st :-

$\green{ɪ = ɪ_{1} +ɪ _{2} } \\ \\ \boxed{ \orange { ɪ = \sf\frac{ - 1}{2(x {}^{2} + 1) } + { \tan}^{ - 1} x + \frac{x}{(x {}^{2} + 1) } + c}}$