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Posted by @Dinesh_ 5N04 5 years, 3 months ago
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Posted by Neha Krishnan 5 years, 3 months ago
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Sia ? 5 years, 3 months ago
As we know,
{tex}a ^ { 3 } + b ^ { 3 } + c ^ { 3 } - 3 a b c ={/tex}{tex}( a + b + c ) \left( a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - a b - b c - c a \right){/tex}
{tex}= ( a + b + c ) \left[ a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - ( a b + b c + c a ) \right]{/tex}
{tex}= 5 \left\{ a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - ( a b + b c + c a ) \right\}{/tex}
{tex}= 5 \left( a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - 10 \right){/tex}
Now, {tex} a + b + c = 5{/tex}
Squaring both sides, we get
{tex}( a + b + c ) ^ { 2 } = 5 ^ { 2 }{/tex}
{tex}\Rightarrow a ^ { 2 } + b ^ { 2 } + c ^ { 2 } + 2 ( a b + b c + c a ) = 25{/tex}
{tex}\therefore a ^ { 2 } + b ^ { 2 } + c ^ { 2 } + 2 ( 10 ) = 25{/tex}
{tex}\Rightarrow a ^ { 2 } + b ^ { 2 } + c ^ { 2 } = 25 - 20 = 5{/tex}
Now, {tex}a ^ {3} + b ^ {3} + c ^ { 3 } - 3 a b c = 5 \left( a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - 10 \right){/tex}
{tex}= 5 ( 5 - 10 ) = 5 ( - 5 ) = - 25{/tex}
Hence, proved.
Posted by Tanushka Rawat 5 years, 4 months ago
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Posted by Sunil Harjai 5 years, 4 months ago
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Sia ? 5 years, 4 months ago
The rational number {tex}\left(\frac{-2}{3}+\frac{1}{4}\right){/tex}{tex}\div{/tex} 2 lies in the middle of the two given rational numbers, that is, {tex}\frac{-2}{3}{/tex} and {tex}\frac{1}{4}{/tex}.
Now, {tex}\left(\frac{-2}{3}+\frac{1}{4}\right){/tex} = {tex}\frac{(-2) \times 4 + 3 \times 1}{12}{/tex} = {tex}\frac{-8+3}{12}{/tex} = {tex}\frac{-5}{12}{/tex}
{tex}\therefore{/tex} {tex}\left(\frac{-2}{3}+\frac{1}{4}\right){/tex} {tex}\div{/tex} 2 = {tex}\frac{-5}{12}{/tex} {tex}\div{/tex} 2 = {tex}\frac{-5}{12}{/tex} {tex}\times{/tex} {tex}\frac{1}{2}{/tex} = {tex}\frac{-5}{24}{/tex}
Thus, the required rational number is {tex}\frac{-5}{24}{/tex}. That is, {tex}\frac{-2}{3}{/tex} < {tex}\frac{-5}{24}{/tex} < {tex}\frac{1}{4}{/tex}.
Posted by Sagar Gupta 5 years, 4 months ago
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Sunny Kumar 5 years, 3 months ago
Narendra Kumar 5 years, 4 months ago
Ruchika Aggarwal 5 years, 4 months ago
Posted by Mahi Singh 5 years, 4 months ago
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Chirag Sharma 5 years, 4 months ago
Manjunatha Setty K A 5 years, 4 months ago
Posted by Ashmit Gardia 5 years, 4 months ago
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Mahi Singh 5 years, 4 months ago
Yogita Ingle 5 years, 4 months ago
Using the identity (a + b)3 = a3 + b3 + 3a2b + 3ab2 , we have , we have
= (3a)3 + (4b)3 + 3 (3a)2(4b) + 3(3a) (4b)2
= 27a3 + 64b3 + 108a2b + 144ab2
which is the required expanded form.
Posted by Sumit Karan 5 years, 4 months ago
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Sia ? 5 years, 4 months ago
According to question it is given that
Radius of cylindrical tub = 12 cm
Depth of cylindrical tub = 20 cm
Let us suppose that (r) be the radius of spherical ball
Again it is given that level of water is raised by 6.75 cm
Now, according to the question,
Volume of spherical ball = Volume of water rise in cylindrical tub
{tex}\Rightarrow \quad \frac { 4 } { 3 } \pi r ^ { 3 } = \pi ( 12 ) ^ { 2 } \times 6.75{/tex}
{tex}\Rightarrow \frac { 4 } { 3 } r ^ { 2 } = 12 \times 12 \times 6. 75{/tex}
{tex}\Rightarrow \quad r ^ { 3 } = \frac { 12 \times 12 \times 6.75 \times 3 } { 4 }{/tex}
{tex}\Rightarrow \quad r ^ { 3 } = 729{/tex}
{tex}\Rightarrow \quad r = \sqrt [ 3 ] { 729 } = 9 \mathrm { cm }{/tex}
Therefore, Radius of the ball = 9 cm
Posted by Yash Sihag 5 years, 4 months ago
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Posted by Hariom Dalal 5 years, 4 months ago
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Sia ? 5 years, 4 months ago
Let in {tex}\bigtriangleup{/tex}ABC, exterior {tex}\angle{/tex}ACE = 115o and {tex}\angle{/tex}A = 35o
We know that,
{tex}\angle{/tex}ACE = {tex}\angle{/tex}A + {tex}\angle{/tex}B (Exterior Angle Theorem)
{tex}\Rightarrow{/tex} 115o = 35o + {tex}\angle{/tex}B
{tex}\angle{/tex}B = 115o – 35o = 80o
Again, in {tex}\bigtriangleup{/tex}ABC,
{tex}\angle{/tex}A + {tex}\angle{/tex}B + {tex}\angle{/tex}C = 180o
(The sum of the three angles of a triangle is 180o)
{tex}\Rightarrow{/tex} 35o + 80o + {tex}\angle{/tex}C = 180o
{tex}\Rightarrow{/tex} 115o + {tex}\angle{/tex}C = 180o
{tex}\Rightarrow{/tex} {tex}\angle{/tex}C = 180o – 115o = 65o
Srishtika Murugan 5 years, 4 months ago
Posted by Balamourougane Balamourougane 5 years, 4 months ago
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Sia ? 5 years, 3 months ago
We have,
{tex}\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}{/tex} = {tex}\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}} \times \frac{3 \sqrt{5}+2 \sqrt{6}}{3 \sqrt{5}+2 \sqrt{6}}{/tex}
= {tex}\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{(3 \sqrt{5}-2 \sqrt{6})(3 \sqrt{5}+2 \sqrt{6})}{/tex}
= {tex}\frac{2 \sqrt{6}(3 \sqrt{5}+2 \sqrt{6})-\sqrt{5}(3 \sqrt{5}+2 \sqrt{6})}{(3 \sqrt{5})^{2}-(2 \sqrt{6})^{2}}{/tex}
= {tex}\frac{6 \sqrt{30}+24-15-2 \sqrt{30}}{45-24}{/tex}
= {tex}\frac{4 \sqrt{30}+9}{21}{/tex}
{tex}\therefore{/tex} {tex}\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}{/tex} = {tex}\frac{4 \sqrt{30}+9}{21}{/tex}
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