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Question Number 27547 by abdo imad last updated on 08/Jan/18 $${let}\:{give}\:{A}=\left(_{\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{1}} ^{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{1}} \right)\:\:\:\:\:{find}\:{A}^{{n}} \:\:\:{and}\:\:{e}^{{A}} \:\:\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 93077 by hmamarques1994@gmail.com last updated on 10/May/20 $$\:\:\mathrm{Se}\:\:\mathrm{f}\left(\sqrt{\mathrm{x}}\:−\mathrm{1}\right)\:=\:\mathrm{x}+\mathrm{6},\:\:\mathrm{log}\left[\mathrm{f}\left(\mathrm{1}\right)\right]\:=\:? \\ $$ Commented by john santu last updated on 11/May/20 $$\mathrm{f}\left(\sqrt{{x}}−\mathrm{1}\right)\:=\:{x}+\mathrm{6}\: \\ $$$${set}\:\sqrt{{x}}\:−\mathrm{1}\:=\:{p}\:\Rightarrow{x}\:=\:\left({p}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${so}\:{f}\left({x}\right)=\:\left({x}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 158598 by HongKing last updated on 06/Nov/21 Commented by MJS_new last updated on 06/Nov/21 $$\mathrm{let}\:{y}={px}\wedge{z}={qx}\:\mathrm{then}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}.\:\mathrm{I}\:\mathrm{get} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\mathrm{6}}\wedge{y}=−\frac{{x}}{\mathrm{3}}\wedge{z}=−\frac{{x}}{\mathrm{2}} \\ $$ Commented by HongKing last…
Question Number 93057 by gonzo last updated on 10/May/20 $${y}={b}\mathrm{1}{x}\mathrm{1}+{b}\mathrm{2}{x}\mathrm{2}+{c} \\ $$$${i}\:{need}\:{to}\:{arrange}\:{the}\:{equation}\:{for}\:{thd}\:{value}\:{of}\:{x}\mathrm{2} \\ $$ Commented by gonzo last updated on 10/May/20 $${i}\:{have}\:{discalcula}\:{and}\:{it}\:{helps}\:{me}\:{work}\:{kt}\:{through} \\ $$ Answered…
Question Number 27507 by NECx last updated on 07/Jan/18 $$\mathrm{2}\sqrt{{x}\:}+{y}=\mathrm{9}….\left(\mathrm{1}\right) \\ $$$${x}+\:\mathrm{2}\sqrt{{y}}=\mathrm{3}….\left(\mathrm{2}\right) \\ $$$$ \\ $$$${solve}\:{the}\:{simultaneous}\:{equation} \\ $$ Commented by Rasheed.Sindhi last updated on 08/Jan/18…
Question Number 27503 by Rasheed.Sindhi last updated on 07/Jan/18 $$\mathrm{If}\:\mathrm{x}=\mathrm{cy}+\mathrm{bz}\:,\mathrm{y}=\mathrm{az}+\mathrm{cx}\:\&\:\mathrm{z}=\mathrm{bx}+\mathrm{ay} \\ $$$$\mathrm{prove}\:\mathrm{that}\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }=\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{1}−\mathrm{b}^{\mathrm{2}} }=\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{1}−\mathrm{c}^{\mathrm{2}} }\:. \\ $$ Answered by mrW1 last updated…
Question Number 93039 by i jagooll last updated on 10/May/20 $$\begin{cases}{{xy}+{yz}\:=\:\mathrm{8}}\\{{yz}+{xz}\:=\:\mathrm{9}}\\{{zx}+{xy}\:=\:\mathrm{5}}\end{cases} \\ $$ Commented by i jagooll last updated on 10/May/20 $$\mathrm{2xy}+\mathrm{2xz}+\mathrm{2yz}\:=\:\mathrm{22} \\ $$$$\mathrm{xy}+\mathrm{xz}+\mathrm{yz}\:=\:\mathrm{11} \\…
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