Question Number 190131 by HeferH last updated on 28/Mar/23 $$\:\mathrm{if}:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{14}\:=\:\mathrm{2}\left(\mathrm{x}\:+\:\mathrm{2y}\:+\:\mathrm{3z}\right) \\ $$$$\:\mathrm{find}:\:\:\mathrm{T}=\frac{\mathrm{xyz}}{\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +\mathrm{z}^{\mathrm{3}} }\: \\ $$ Answered by som(math1967) last updated…
Question Number 59053 by logan last updated on 04/May/19 $${f}={v} \\ $$$${v}=\mathrm{47}×\mathrm{48} \\ $$$${f}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 190115 by HeferH last updated on 27/Mar/23 $$\mathrm{if}:\:\:\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}+\mathrm{1}\right)\:=\:\mathrm{b} \\ $$$$\:\mathrm{find}:\:\:\mathrm{P}\:=\:\:\sqrt{\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} −\mathrm{3ab}} \\ $$ Answered by som(math1967) last updated on 27/Mar/23 $$\left({a}+{b}\right)\left({a}+\mathrm{1}\right)={b} \\…
Question Number 124567 by mr W last updated on 04/Dec/20 $${find}\:{the}\:{smallest}\:{integer}\:{which}\:{has} \\ $$$$\mathrm{28}\:{divisors}\:{and}\:{is}\:{divisible}\:{by}\:\mathrm{28}. \\ $$ Answered by mr W last updated on 05/Dec/20 $${say}\:{the}\:{number}\:{is}\:{n}={p}^{{a}} {q}^{{b}}…
Question Number 190093 by mnjuly1970 last updated on 27/Mar/23 $$ \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$ \\ $$$${c}=\:\left(\:\sqrt{\mathrm{5}}\:+\mathrm{2}\right)^{\:\frac{\mathrm{1}}{\mathrm{3}}} \:−\:\left(\sqrt{\mathrm{5}}\:−\mathrm{2}\right)^{\:\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{is}\:\:\:\mathrm{a}\:\:{rational}\:\:\mathrm{number}. \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:…
Question Number 190091 by Shrinava last updated on 27/Mar/23 $$\mathrm{1}.\:\mathrm{Find}\:\:\:\mathrm{sin52}°\:+\:\mathrm{sin8}°\:−\:\mathrm{cos22}° \\ $$$$\mathrm{2}.\:\mathrm{If}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\:=\:\mathrm{6}\:\:\:\mathrm{find}\:\:\:\mathrm{a}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} } \\ $$$$\mathrm{3}.\:\mathrm{Find}\:\:\:\frac{\mathrm{tan32}°\:+\:\mathrm{tan13}°}{\mathrm{1}\:−\:\mathrm{tan32}°\:\centerdot\:\mathrm{tan13}°} \\ $$ Answered by BaliramKumar last updated…
Question Number 190076 by Spillover last updated on 26/Mar/23 Answered by som(math1967) last updated on 27/Mar/23 $${a}.\:\:\:{f}^{\boldsymbol{{l}}} \left({x}\right)=\:\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\:\frac{\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)}−\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{sinx}}}}{\boldsymbol{{h}}} \\ $$$$\:=\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{sinx}}−\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)}{\boldsymbol{{h}}\left\{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)\right\}\left(\mathrm{1}+\boldsymbol{{sinx}}\right)} \\ $$$$=\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{−\mathrm{2}{sin}\left(\frac{{h}}{\mathrm{2}}\right){cos}\left({x}+\frac{{h}}{\mathrm{2}}\right)}{{h}\left(\mathrm{1}+\boldsymbol{{sinx}}\right)\left\{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)\right\}}…
Question Number 190073 by Spillover last updated on 26/Mar/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 190075 by Spillover last updated on 26/Mar/23 Answered by PowerMaths last updated on 27/Mar/23 $$\left({a}\right)\:{p}_{\mathrm{1}} =\:\bot^{{r}} \:{from}\:{A}\left(\mathrm{1},\mathrm{3}\right)\:{to}\:{line}\:=\frac{\mid\mathrm{3}×\mathrm{1}+\mathrm{4}×\mathrm{3}−\mathrm{9}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }}\:=\:\frac{\mathrm{6}}{\mathrm{5}}\:=\:\mathrm{1}.\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:{p}_{\mathrm{2}} \:=\:\bot^{{r}} \:{from}\:{B}\left(\mathrm{2},\mathrm{7}\right)\:{to}\:{line}\:=\frac{\mid\mathrm{3}×\mathrm{2}+\mathrm{4}×\mathrm{7}−\mathrm{9}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}}…
Question Number 124522 by bemath last updated on 03/Dec/20 Answered by mr W last updated on 04/Dec/20 $${CD}=\mathrm{25}−\mathrm{16}=\mathrm{9} \\ $$$${AC}^{\mathrm{2}} ={BC}×{CD}=\mathrm{16}×\mathrm{9} \\ $$$${AC}=\mathrm{4}×\mathrm{3}=\mathrm{12} \\ $$$${AB}^{\mathrm{2}}…