Question Number 170435 by MathsFan last updated on 23/May/22 $$\boldsymbol{{A}}\:\boldsymbol{{line}}\:\boldsymbol{{is}}\:\boldsymbol{{formed}}\:\boldsymbol{{by}}\:\boldsymbol{{joining}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{points}}\:\boldsymbol{{A}}\left(\mathrm{7},\mathrm{0}\right)\:\boldsymbol{{and}}\:\boldsymbol{{B}}\left(\mathrm{0},\mathrm{2}\right).\:\boldsymbol{{Obtain}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{equation}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{straight}}\:\boldsymbol{{line}} \\ $$$$\boldsymbol{{joining}}\:\boldsymbol{{AC}}\:\boldsymbol{{such}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{x}}−\boldsymbol{{axis}} \\ $$$$\boldsymbol{{bisects}}\:\boldsymbol{{the}}\:\boldsymbol{{angle}}\:\boldsymbol{{BAC}}. \\ $$ Answered by MikeH last updated…
Question Number 170426 by daus last updated on 23/May/22 Answered by Rasheed.Sindhi last updated on 23/May/22 $$\frac{\mathrm{1}}{\mathrm{3}.\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}.\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{7}.\mathrm{9}}+… \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)}=\frac{{A}}{\mathrm{2}{k}+\mathrm{1}}+\frac{{B}}{\mathrm{2}{k}+\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:{A}\left(\mathrm{2}{k}+\mathrm{3}\right)+{B}\left(\mathrm{2}{k}+\mathrm{1}\right)=\mathrm{1}…
Question Number 104876 by mathocean1 last updated on 24/Jul/20 $${A}\:{ballot}\:{box}\:{contains}\:\mathrm{7}\:{balls}\left(\mathrm{3}\:{are}\:\right. \\ $$$$\left.{black}\right).\:{we}\:{draw}\:{successively}\:{and} \\ $$$${reputing}\left({inside}\right)\:\mathrm{5}\:{balls}. \\ $$$${What}\:{is}\:{the}\:{number}\:{of}\:{possibilities}\: \\ $$$${to}\:{have}\:{one}\:{black}\:{ball}? \\ $$ Terms of Service Privacy Policy…
Question Number 104872 by mathocean1 last updated on 24/Jul/20 $${solve}\:{for}\:{a}\:{and}\:{b}: \\ $$$$\begin{cases}{{a}+{b}=\mathrm{60}°}\\{{tana}=\sqrt{\mathrm{2}}{tanb}}\end{cases} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170380 by daus last updated on 22/May/22 Answered by amin96 last updated on 22/May/22 $${picagor}\:{numbers} \\ $$$$\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \:\:\:\:\:\: \\ $$ Commented…
Question Number 39312 by kunal1234523 last updated on 05/Jul/18 $${prove}\:{that} \\ $$$$\left({tan}\:\mathrm{4}{a}+{tan}\:\mathrm{2}{a}\right)\left(\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{3}{a}\:{tan}^{\mathrm{2}} {a}\right)=\mathrm{2}{tan}\:\mathrm{3}{a}\:{sec}^{\mathrm{2}} {a} \\ $$ Answered by kunal1234523 last updated on 05/Jul/18 Commented…
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Question Number 170350 by SANOGO last updated on 21/May/22 $${developpement}\:{limite}\:{de}\:{ordre}\:\mathrm{3} \\ $$$${sin}^{\mathrm{2}} {x}\:\:\:\:{et}\:\:\:{cos}^{\mathrm{2}} {x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 104799 by mathocean1 last updated on 23/Jul/20 Commented by mathocean1 last updated on 23/Jul/20 $${I}\:{is}\:{the}\:{point}\:{of}\:{intersection}\:{of}\:\left({AC}\right)\:\: \\ $$$${and}\:\left({BE}\right). \\ $$$${Given}\:{AC}=\mathrm{10}\sqrt{\mathrm{3}}\:{m}\:\:{and}\:{AB}=\mathrm{10}{m}. \\ $$$$\left.\mathrm{1}\right)\:{Determinate}\:{AI}\:{in}\:{each}\:{case}: \\ $$$$\left.{a}\right)\:{a}=\mathrm{2}{b}…