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Author: Tinku Tara

From-a-poin-P-outside-a-circle-a-tangent-is-drawn-to-a-circle-at-T-from-P-a-line-is-drawn-to-circle-at-A-and-B-Find-lenth-PT-giving-that-i-PA-6-AB-8-ii-PB-18-PT-12-

Question Number 137946 by otchereabdullai@gmail.com last updated on 08/Apr/21 $$\mathrm{From}\:\mathrm{a}\:\mathrm{poin}\:\mathrm{P}\:\mathrm{outside}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{a} \\ $$$$\mathrm{tangent}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{at}\:\mathrm{T}. \\ $$$$\mathrm{from}\:\mathrm{P}\:\mathrm{a}\:\mathrm{line}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to}\:\mathrm{circle}\:\mathrm{at} \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:.\:\mathrm{Find}\:\mathrm{lenth}\:\mathrm{PT}\:\mathrm{giving}\:\mathrm{that} \\ $$$$\mathrm{i}.\:\mathrm{PA}\:=\mathrm{6} \\ $$$$\:\:\:\:\mathrm{AB}=\mathrm{8} \\ $$$$\mathrm{ii}.\:\mathrm{PB}=\mathrm{18} \\ $$$$\:\:\:\:\:\mathrm{PT}=\mathrm{12} \\…

Eliminate-if-x-a-sec-3-and-y-b-tan-3-

Question Number 6872 by hanandmishra026@gmail.com last updated on 31/Jul/16 $${Eliminate}\:\theta\:\:{if}\:{x}={a}\:\mathrm{sec}^{\mathrm{3}} \theta\:{and}\:\:{y}={b}\:\mathrm{tan}^{\mathrm{3}} \theta \\ $$ Commented by Yozzii last updated on 31/Jul/16 $${sec}\theta=\left(\frac{{x}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} ,\:{tan}\theta=\left(\frac{{y}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${sec}^{\mathrm{2}}…

0-x-1-2-x-1-ln-2-x-1-dx-pi-2ln-2-2-

Question Number 137941 by Ñï= last updated on 08/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}^{{x}} −\mathrm{1}}{ln}\:\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}{dx}=\frac{\pi}{\mathrm{2}{ln}^{\mathrm{2}} \mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Determine-the-term-independent-of-x-in-the-expansion-x-1-x-2-3-x-1-3-1-x-1-x-x-1-2-10-

Question Number 137943 by john_santu last updated on 08/Apr/21 $${Determine}\:{the}\:{term}\:{independent} \\ $$$${of}\:{x}\:{in}\:{the}\:{expansion}\: \\ $$$$\:\:\:\:\left(\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}/\mathrm{3}} −{x}^{\mathrm{1}/\mathrm{3}} +\mathrm{1}}\:−\frac{{x}−\mathrm{1}}{{x}−{x}^{\mathrm{1}/\mathrm{2}} }\:\right)^{\mathrm{10}} \:. \\ $$ Answered by EDWIN88 last updated…

ln-x-x-2-a-2-2-dx-

Question Number 137937 by benjo_mathlover last updated on 08/Apr/21 $$\:\underset{−\infty} {\overset{\:\:\:\infty} {\int}}\frac{\mathrm{ln}\left(\:\mid{x}\mid\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:=? \\ $$ Answered by Ñï= last updated on 14/Apr/21 $$\int_{−\infty}…

n-0-4n-2n-1-16-15-3-27-pi-2-5-25-ln-1-5-2-

Question Number 137936 by Ñï= last updated on 08/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{4}{n}}\\{\mathrm{2}{n}}\end{pmatrix}^{−\mathrm{1}} =\frac{\mathrm{16}}{\mathrm{15}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{27}}\pi−\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{25}}{ln}\left(\mathrm{1}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$ Commented by Dwaipayan Shikari last updated on 08/Apr/21 $$\underset{{n}=\mathrm{0}} {\overset{\infty}…

0-1-0-1-ln-1-x-y-dxdy-5-2-ln2-1-2-lnpi-9-4-

Question Number 137939 by Ñï= last updated on 08/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {ln}\Gamma\left(\mathrm{1}+{x}+{y}\right){dxdy}=\frac{\mathrm{5}}{\mathrm{2}}{ln}\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\pi−\frac{\mathrm{9}}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com