Question Number 9323 by uchechukwu okorie favour last updated on 30/Nov/16 $${if}\:\:{x}=\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}}\:{make}\:{r}\:{the}\: \\ $$$${subject}\:{of}\:{the}\:{formula} \\ $$ Commented by geovane10math last updated on 30/Nov/16 $${x}\:=\:\frac{{a}\left(−{r}^{{n}}…
Question Number 9312 by tawakalitu last updated on 29/Nov/16 $$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{xy}\:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{23}\:\:\:\:…….\:\left(\mathrm{i}\right) \\ $$$$\mathrm{xz}\:+\:\mathrm{x}\:+\:\mathrm{z}\:=\:\mathrm{41}\:\:\:\:……..\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{yz}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{27}\:\:\:\:\:……..\:\left(\mathrm{iii}\right) \\ $$ Commented by RasheedSoomro last updated on 30/Nov/16…
Question Number 9278 by j.masanja06@gmail.com last updated on 28/Nov/16 $$\mathrm{represent}\:\mathrm{in}\:\mathrm{sigma}\:\mathrm{notation}\: \\ $$$$−\mathrm{1}+\mathrm{4}−\mathrm{9}+\mathrm{16}……………… \\ $$ Commented by FilupSmith last updated on 28/Nov/16 $$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {n}^{\mathrm{2}}…
Question Number 9279 by j.masanja06@gmail.com last updated on 28/Nov/16 $$\mathrm{evalute}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\overset{\mathrm{5}} {\sum}_{\mathrm{m}=\mathrm{2}\:} \mathrm{m}^{\mathrm{4}} \\ $$ Answered by mrW last updated on 28/Nov/16 $$\mathrm{please}\:\mathrm{refer}\:\mathrm{to}\:\mathrm{Q9232}\:\mathrm{and}\:\mathrm{Q9265}. \\…
Question Number 140346 by mathsuji last updated on 06/May/21 $${if},\:{x}>\mathrm{4}\:;\:\mathrm{5}!\centerdot{x}!={z}!\:;\:{find}:\:{x}+{z}=? \\ $$ Answered by hknkrc46 last updated on 06/May/21 $$\mathrm{5}!\:=\:\frac{\boldsymbol{{z}}!}{\boldsymbol{{x}}!}\:\Rightarrow\:\begin{cases}{\boldsymbol{{x}}\:=\:\mathrm{5}!\:−\:\mathrm{1}\:\Rightarrow\:\boldsymbol{{z}}\:=\:\mathrm{120}}\\{\boldsymbol{{x}}\:+\:\boldsymbol{{z}}\:=\:\mathrm{119}\:+\:\mathrm{120}\:=\:\mathrm{239}}\end{cases} \\ $$ Commented by mathsuji…
Question Number 74801 by behi83417@gmail.com last updated on 30/Nov/19 $$\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} =\mathrm{23}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{32}}\end{cases}\:\:\:\:\:\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}}\:. \\ $$ Commented by MJS last updated on 30/Nov/19 $$\mathrm{if}\:\mathrm{you}\:\mathrm{plot}\:\mathrm{these}\:\mathrm{you}\:\mathrm{see}\:\mathrm{there}'\mathrm{s}\:\mathrm{only}\:\mathrm{one} \\…
Question Number 74786 by chess1 last updated on 30/Nov/19 Commented by abdomathmax last updated on 30/Nov/19 $${S}\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{2019}} \:{k}\left(\frac{\mathrm{1}}{\mathrm{2020}}\right)^{{k}} \:={w}\left(\frac{\mathrm{1}}{\mathrm{2020}}\right)\:{with} \\ $$$${w}\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{\mathrm{2019}} \:{kx}^{{k}} \:\:\:{we}\:{have}\:{for}\:{x}\neq\mathrm{1}…
Question Number 9236 by tawakalitu last updated on 24/Nov/16 $$\mathrm{Solve}\:\:\mathrm{simultaneously} \\ $$$$\frac{\mathrm{x}}{\mathrm{y}\:+\:\mathrm{1}_{\:} }\:+\:\frac{\mathrm{y}}{\mathrm{x}\:+\:\mathrm{1}}\:=\:\frac{\mathrm{5}}{\mathrm{3}}\:\:\:\:………….\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{2}\:\:\:\:\:\:\:\:………..\:\left(\mathrm{ii}\right) \\ $$ Answered by RasheedSoomro last updated on…
Question Number 9229 by tawakalitu last updated on 24/Nov/16 $$\mathrm{2}^{\mathrm{3x}\:+\:\mathrm{1}} \:−\:\mathrm{3}.\mathrm{2}^{\mathrm{2x}} \:+\:\mathrm{2}^{\mathrm{x}\:+\:\mathrm{1}} \:=\:\mathrm{2x} \\ $$$$\mathrm{Find}\:\mathrm{x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 9219 by tawakalitu last updated on 23/Nov/16 $$\mathrm{Show}\:\mathrm{that}:\: \\ $$$$\left(\mathrm{a}\:+\:\mathrm{b}\right)\left[\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }\right]\left[\frac{\mathrm{a}^{\mathrm{4}} }{\mathrm{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{b}^{\mathrm{4}} }{\mathrm{a}^{\mathrm{2}} }\right]\:\geqslant\:\mathrm{8}\sqrt{\mathrm{ab}} \\ $$ Commented by Yozzias last updated…