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Category: Mensuration

Question-37989

Question Number 37989 by ajfour last updated on 20/Jun/18 Commented by ajfour last updated on 20/Jun/18 $$\left({i}\right){Find}\:{area}\:{of}\:{blue}\:{square}\:{and}\: \\ $$$${red}\:{square}. \\ $$$$\left({ii}\right){Find}\:\theta\:{for}\:{which}\:{corners}\:{of} \\ $$$${red}\:{square}\:{lies}\:{on}\:{sides}\:{of}\:{blue} \\ $$$${square}.…

f-x-y-z-3x-2-y-x-3-y-3-2z-prove-that-the-function-has-a-potential-to-be-determined-

Question Number 168979 by MikeH last updated on 22/Apr/22 $${f}\left({x},{y},{z}\right)\:=\:\left(\mathrm{3}{x}^{\mathrm{2}} {y},{x}^{\mathrm{3}} +{y}^{\mathrm{3}} ,\:\mathrm{2}{z}\right) \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{has}\:\mathrm{a}\:\mathrm{potential} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{determined}. \\ $$ Terms of Service Privacy Policy Contact:…

Question-168906

Question Number 168906 by cortano1 last updated on 21/Apr/22 Answered by mr W last updated on 21/Apr/22 $$\mathrm{tan}\:\frac{{B}}{\mathrm{2}}=\frac{\frac{{b}}{{c}}}{\mathrm{1}+\frac{{a}}{{c}}}=\frac{{b}}{{a}+{c}}=\frac{{b}}{{a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\frac{{A}}{\mathrm{2}}=\frac{\frac{{a}}{{c}}}{\mathrm{1}+\frac{{b}}{{c}}}=\frac{{a}}{{b}+{c}}=\frac{{a}}{{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\…

Question-103171

Question Number 103171 by bemath last updated on 13/Jul/20 Commented by bobhans last updated on 13/Jul/20 $$\left(\mathrm{2}{a}\right)\:{a}\:{sector}\:{of}\:{a}\:{circle}\:{with}\:{radius}\:{r}\:{has} \\ $$$${perimeter}\:\mathrm{20}\:{units}\:{and}\:{area}\:\mathrm{21}\:{square} \\ $$$${units}\:.\:{Find}\:{all}\:{possible}\:{of}\:{r}\:{and}\:{the}\: \\ $$$${corresponding}\:{value}\:{of}\:\theta\:. \\ $$…

Question-103001

Question Number 103001 by I want to learn more last updated on 12/Jul/20 Answered by ajfour last updated on 12/Jul/20 $${let}\:{new}\:{depth}\:{be}\:\boldsymbol{{y}}. \\ $$$$\boldsymbol{{V}}=\pi{R}^{\mathrm{2}} {H}=\pi{R}^{\mathrm{2}} \left({y}−\frac{{h}}{\mathrm{2}}\right)+\pi\left(\frac{{h}}{\mathrm{2}}\right)\left({R}^{\mathrm{2}}…

1-m-1-n-tan-2-mpi-2n-1-2-m-1-n-tan-2-mpi-2n-1-

Question Number 102910 by bramlex last updated on 11/Jul/20 $$\left(\mathrm{1}\right)\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{m}\pi}{\mathrm{2n}+\mathrm{1}}\right) \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{m}\pi}{\mathrm{2n}+\mathrm{1}}\right)\: \\ $$ Answered by floor(10²Eta[1]) last updated…