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Preeti Dabral 1 year, 11 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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Preeti Dabral 1 year, 11 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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Preeti Dabral 1 year, 11 months ago
{tex}\text { Let } \alpha \text { and } \beta \text { are the zeroes of the polynomial } f(x)=a x^2+b x+c \text {. }{/tex}
{tex}\therefore \quad(\alpha+\beta)=\frac{-\mathbf{b}}{\mathbf{a}} and \alpha \beta=\frac{\mathbf{c}}{\mathbf{a}} According to the question, \frac{1}{\alpha} and \frac{1}{\beta} are the zeroes of the required quadratic polynomial{/tex}
Sum of zeroes of required polynomial
{tex} \begin{aligned} \mathbf{S}^{\prime} & =\frac{1}{\alpha}+\frac{1}{\beta} \\ & =\frac{\alpha+\beta}{\alpha \beta} \\ & =\frac{-\mathbf{b}}{\mathbf{c}} \end{aligned} [From equation (i) and (ii)] and product of zeroes of required polynomial =\frac{1}{\alpha} \times \frac{1}{\beta}. P^{\prime}=\frac{1}{\alpha \beta} =\frac{a}{c} [From equation (ii)]{/tex}
Equation of the required polynomial.
{tex} =\mathbf{k}\left(\mathbf{x}^2-\mathbf{S}^{\prime} \mathbf{x}+\mathbf{p}^{\prime}\right) where $k$ is any non-zero constant =k\left(x^2-\left(\frac{-b}{c}\right) x+\frac{a}{c}\right) [From equation (iii) and (iv)] =k\left(x^2+\frac{b}{c} x+\frac{a}{c}\right) {/tex}
Posted by Shanmuga Kani 1 year, 11 months ago
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Preeti Dabral 1 year, 11 months ago
The given equations are
{tex}71x + 37y = 253{/tex}.........(i)
{tex}37x + 71y = 287{/tex}.........(ii)
Adding (i) and (ii),we get
{tex}108x + 108y = 540{/tex}
{tex}108(x + y) = 540{/tex}
{tex}\therefore x + y = \frac { 540 } { 108 } = 5{/tex}......(iii)
Subtracting (ii) from (i),
{tex}34x - 34y = 253 - 287 = -34{/tex}
{tex}34(x - y) = -34{/tex}
{tex}\therefore x - y = - \frac { 34 } { 34 } = - 1{/tex}....(iv)
Adding (iii) and (iv)
2x = 5 - 1 = 4
{tex}\Rightarrow x = 2{/tex}
Subtracting (iv) from (iii)
2y = 5 + 1 = 6
{tex}\Rightarrow y = 3{/tex}
{tex}\therefore{/tex} Solution is x = 2, y = 3
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Kartik Upreti 1 year, 10 months ago
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