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Author: Tinku Tara

Find-A-1-2-3-323-

Question Number 213413 by hardmath last updated on 04/Nov/24 $$\mathrm{Find}: \\ $$$$\mathrm{A}=\:\left[\sqrt{\mathrm{1}}\right]\:+\:\left[\sqrt{\mathrm{2}}\right]\:+\:\left[\sqrt{\mathrm{3}}\:\right]+…+\:\left[\sqrt{\mathrm{323}}\right]\:=\:? \\ $$ Answered by mehdee7396 last updated on 04/Nov/24 $$=\left(\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+\right)+\left(\left[\mathrm{4}\right]+\left[\mathrm{5}\right]+…+\left[\left[\sqrt{\mathrm{8}}\right]\right)\right. \\ $$$$+\left(\left[\sqrt{\mathrm{9}}\right]+\left[\sqrt{\mathrm{10}}\right]+…+\left[\sqrt{\mathrm{24}}\right]\right)+… \\…

One-simple-Equation-pls-prove-this-property-j-1-N-a-j-k-1-M-b-k-j-1-N-k-1-M-a-j-b-k-and-j-0-N-f-a-b-a-N-j-b-a-N-k-0-M-g-a-b-a-M-k-b-a-M-j-

Question Number 213398 by issac last updated on 04/Nov/24 $$\mathrm{One}\:\mathrm{simple}\:\mathrm{Equation} \\ $$$$\mathrm{pls}\:\mathrm{prove}\:\mathrm{this}\:\mathrm{property} \\ $$$$\underset{{j}=\mathrm{1}} {\overset{{N}} {\sum}}\:{a}_{{j}} \centerdot\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}{b}_{{k}} =\underset{{j}=\mathrm{1}} {\overset{{N}} {\sum}}\centerdot\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}\:{a}_{{j}} {b}_{{k}}…

Find-1-2-5-1-4-5-3-5-1-4-2-

Question Number 213374 by hardmath last updated on 03/Nov/24 $$\mathrm{Find}:\:\:\:\frac{\mathrm{1}\:−\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\:+\:\sqrt{\mathrm{5}}}{\left(\sqrt{\mathrm{3}}\:−\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\right)^{\mathrm{2}} }\:=\:? \\ $$ Commented by Frix last updated on 03/Nov/24 $$\frac{\mathrm{29}−\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{8}−\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{5}^{\frac{\mathrm{3}}{\mathrm{4}}} +\frac{\mathrm{13}−\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{\mathrm{18}−\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$…

Question-213342

Question Number 213342 by Spillover last updated on 03/Nov/24 Commented by MrGaster last updated on 03/Nov/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{\:\:\int{H}_{{x}} ^{\sqrt{\pi}} {dx}=\frac{{H}_{{x}} ^{\sqrt{\pi}+\mathrm{1}} }{\:\sqrt{\pi}+\mathrm{1}}+{C}\:\:}\\\hline\end{array} \\ $$ Terms of…

Question-213343

Question Number 213343 by Spillover last updated on 03/Nov/24 Answered by MrGaster last updated on 03/Nov/24 $$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\left({x}\right)\mathrm{sin}\left({y}\right)\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}\left({z}\right)}{{x}+{y}+{z}}{dx}\right){dydx} \\ $$$$=\int_{\mathrm{0}}…