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Sia ? 6 years, 6 months ago
We have to express the given decimal in fractional form. So for that, let {tex}x = 0.3 \overline { 65 }{/tex}, then
x = 0.3656565.... ...(i)
10x = 3.656565.... ...(ii)
1000x = 365.656565.... ....(iii)
Subtracting (ii) from (iii), we obtain
{tex}990x = 362 \\ \Rightarrow x = \frac { 362 } { 990 } = \frac { 181 } { 495 }{/tex}
Posted by Deepanshu Thakue 7 years, 8 months ago
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Posted by Rachel Thomas 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
{tex}\begin{array}{l}(2\times5)^{\mathrm n}=2^{\mathrm n}\times5^{\mathrm n}\end{array}{/tex}
{tex}\text{=10}^n{/tex}
{tex}\text{If n=0 then 10}^0\text{=1}{/tex}
{tex}\text{If n>0 then 10}^n\text{ will end with 0 }{/tex}
{tex}\mathrm{If}\;\mathrm n<0\;\mathrm{then}\;10^{\mathrm n}\;\mathrm{ends}\;\mathrm{with}1\;(\mathrm e.\mathrm g.\;0.1,0.01,0.001){/tex}
Hence for all values of n, {tex}2^n\times 5^n{/tex} can never end with 5.
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Posted by Prajval Rawat 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
{tex}\angle{/tex}OAP = 90o ...........(1) [Angle between tangent and radius through the point of contact is 90o ]
{tex}\angle{/tex}OBP = 90o ...........(2) [Angle between tangent and radius through the point of contact is 90o ]
{tex}\therefore{/tex} OAPB is quadrilateral
{tex}\therefore{/tex} {tex}\angle{/tex}APB + {tex}\angle{/tex}AOB + {tex}\angle{/tex}OAP + {tex}\angle{/tex}OBP = 360o [Angle sum property of a quadrilateral]
{tex}\Rightarrow{/tex} {tex}\angle{/tex}APB + {tex}\angle{/tex}AOB + 90o + 90o = 360o [From (1) and (2)]
{tex}\Rightarrow{/tex} {tex}\angle{/tex}APB + {tex}\angle{/tex}AOB = 180o
{tex}\Rightarrow{/tex} {tex}\angle{/tex}APB and {tex}\angle{/tex}AOB are supplementary
Posted by Prajval Rawat 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
For Hemisphere,
Radius(r) = 1 cm
{tex}\therefore {/tex} Volume {tex}= \frac { 2 } { 3 } \pi r ^ { 3 }{/tex}
{tex}= \frac { 2 } { 3 } \pi ( 1 ) ^ { 3 }{/tex}
{tex}= \frac { 2 } { 3 } \pi \mathrm { cm } ^ { 3 }{/tex}
For cone,
Radius of the base(r) = 1 cm
Height (h) = 1 cm
{tex}\therefore {/tex} Volume{tex}= \frac { 1 } { 3 } \pi r ^ { 2 } h{/tex}
{tex}= \frac { 1 } { 3 } \pi ( 1 ) ^ { 2 } ( 1 ) = \frac { 1 } { 3 } \pi \operatorname { cm } ^ { 3 }{/tex}
Therefore, volume of the solid
=volume of the hemisphere + volume of cone
{tex}= \frac { 2 } { 3 } \pi + \frac { 1 } { 3 } \pi = \pi \mathrm { cm } ^ { 3 }{/tex}
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Abhishek Reddy 7 years, 8 months ago
1Thank You