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$$2\frac{1.5}{5}$$ + 2$$\frac{1}{6}$$ - $$1\frac{3.5}{15}$$ = $$\left( {\frac{{{{(?)}^{\frac{1}{3}}}}}{4}} \right)$$  + $$1\frac{7}{30}$$

$$2\frac{1.5}{5}$$ + 2$$\frac{1}{6}$$ - $$1\frac{3.5}{15}$$ = $$\left( {\frac{{{{(?)}^{\frac{1}{3}}}}}{4}} \right)$$  + $$1\frac{7}{30}$$ Correct Answer 512

$$\eqalign{ & 2\frac{{1.5}}{5} + 2\frac{1}{6} - 1\frac{{3.5}}{{15}} = \frac{{{x^{\frac{1}{3}}}}}{4} + 1\frac{7}{{30}} \cr & \Rightarrow \frac{{11.5}}{5} + \frac{{13}}{6} - \frac{{18.5}}{{15}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr} $$
⇒ L.C.M. of 5, 6 and 15 is 30
$$\eqalign{ & \Rightarrow \frac{{69 + 65 - 37}}{{30}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr & \Rightarrow \frac{{97}}{{30}} = \frac{{{x^{\frac{1}{3}}}}}{4} + \frac{{37}}{{30}} \cr & \Rightarrow \frac{{{x^{\frac{1}{3}}}}}{4} = \frac{{97}}{{30}} - \frac{{37}}{{30}} \cr & \Rightarrow {x^{\frac{1}{3}}} = \frac{{60}}{{30}} \times 4 \cr & \Rightarrow {x^{\frac{1}{3}}} = 8 \cr & \Rightarrow x = {\left( 8 \right)^3} \cr & \Rightarrow x = 512 \cr} $$
Hence, the number is 512

Related Questions

$$\frac{{\frac{1}{3}.\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4}.\frac{1}{4} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{5}.\frac{1}{5}}}{{\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4} + \frac{1}{5}.\frac{1}{5} - \left( {\frac{1}{3}.\frac{1}{4} + \frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{3}} \right)}}{\text{ is?}}$$
The simplified value of $$\frac{{\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right)\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)\left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)}}{{\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) + \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)}} = ?$$
If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
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Let the function
\[{\text{f}}\left( \theta \right) = \left| {\begin{array}{*{20}{c}} {\sin \theta }&{\cos \theta }&{\tan \theta } \\ {\sin \left( {\frac{\pi }{6}} \right)}&{\cos \left( {\frac{\pi }{6}} \right)}&{\tan \left( {\frac{\pi }{6}} \right)} \\ {\sin \left( {\frac{\pi }{3}} \right)}&{\cos \left( {\frac{\pi }{3}} \right)}&{\tan \left( {\frac{\pi }{3}} \right)} \end{array}} \right|\
The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$   = ?
$$\frac{{{{\left( {4.53 - 3.07} \right)}^2}}}{{\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}} + \, $$     $$\frac{{{{\left( {3.07 - 2.15} \right)}^2}}}{{\left( {2.15 - 4.53} \right)\left( {4.53 - 3.07} \right)}} + \,\, $$     $$\frac{{{{\left( {2.15 - 4.53} \right)}^2}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)}}$$     is simplified to :
Consider the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}\left( {\text{t}} \right)}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}\left( {\text{t}} \right)}}{{{\text{dt}}}} + {\text{y}}\left( {\text{t}} \right) = \delta \left( {\text{t}} \right)$$      with $${\left. {{\text{y}}\left( {\text{t}} \right)} \right|_{{\text{t}} = 0}} = - 2$$   and $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}} = 0.$$
The numerical value of $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}}$$   is
What is the simplified value of \(\left( {{x^{32}} + \frac{1}{{{x^{32}}}}} \right)\left( {{x^8} + \frac{1}{{{x^8}}}} \right)\left( {x - \frac{1}{x}} \right)\left( {{x^{16}} + \frac{1}{{{x^{16}}}}} \right)\left( {x + \frac{1}{x}} \right)\left( {{x^4} + \frac{1}{{{x^4}}}} \right)?\)