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Preeti Dabral 2 years, 10 months ago
According to the question,
Radius of the cones= r
Height of the upper cone= h
Height of the lower cone= H
Volume of the upper cone {tex}= \frac { 1 } { 3 } \pi r ^ { 2 } h{/tex}
Volume of the lower cone {tex}= \frac { 1 } { 3 } \pi r ^ { 2 } H{/tex}
Total volume of both the cones {tex}= \frac { 1 } { 3 } \pi r ^ { 2 } h + \frac { 1 } { 3 } \pi r ^ { 2 } H{/tex}
{tex}= \frac { 1 } { 3 } \pi r ^ { 2 } ( h + H ){/tex}
The quantity of water displaced by the cones = The total volume of both the cones
therefore,
The quantity of water displaced will be {tex}\frac { 1 } { 3 } \pi r ^ { 2 } ( h + H ) \text { units } ^ { 3 }{/tex}
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Posted by Rdm 😈 2 years, 10 months ago
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Preeti Dabral 2 years, 10 months ago
{tex}Here, we have
\begin{aligned}
& \Rightarrow(x-3)(x-4)=34 / 33^2 \\
& \Rightarrow x^2-7 x+12-34 / 33^2=0 \\
& \Rightarrow x^2-7 x+13034 / 33^2=0 \\
& \Rightarrow x^2-7 x+98 / 33 \times 133 / 33=0 \\
& \Rightarrow x^2-231 / 33 x+98 / 33 \times 133 / 33=0
\end{aligned}
{/tex}
{tex}By using factorization method, we get
\begin{aligned}
& \Rightarrow x^2-(98 / 33+133 / 33) x+98 / 33 \times 133 / 33=0 \\
& \Rightarrow x^2-98 / 33 x-133 / 33 x+98 / 33 \times 133 / 33=0 \\
& \Rightarrow\left(x^2-98 / 33 x\right)-(133 / 33 x-98 / 33 \times 133 / 33)=0 \\
& \Rightarrow x(x-98 / 33)-133 / 33(x-98 / 33)=0 \\
& \Rightarrow(x-98 / 33)(x-133 / 33)=0 \\
& \Rightarrow x-98 / 33=0 \text { or } x-133 / 33= \\
& \Rightarrow x=98 / 33,133 / 33 \\
& \text { Hence, } x=x=98 / 33,133 / 33
\end{aligned}
{/tex}
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Aarti Mahato 2 years, 10 months ago
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