Question Number 158101 by zainaltanjung last updated on 31/Oct/21 $$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{image}\:\mathrm{of}\:\mathrm{points}\: \\ $$$$\mathrm{K}\:\left(\mathrm{5},\mathrm{2}\right)\:\mathrm{and}\:\mathrm{L}\:\left(\mathrm{1}\:\mathrm{5}\right)\:\mathrm{after}\: \\ $$$$\mathrm{being}\:\mathrm{reflected}\:\mathrm{about}\:\mathrm{the}\:\mathrm{x} \\ $$$$\mathrm{axis}. \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 158102 by zainaltanjung last updated on 31/Oct/21 $$ \\ $$$$\mathrm{Given}\:\mathrm{the}\:\mathrm{quadrilateral}. \\ $$$$\mathrm{D}.\mathrm{ABC}\:\mathrm{Find}\:\mathrm{the}\:\mathrm{measure}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{CD}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 158069 by zainaltanjung last updated on 30/Oct/21 Answered by Rasheed.Sindhi last updated on 30/Oct/21 $${Let}\:{the}\:{side}\:{of}\:\:{old}\:\:{hexagon}={a} \\ $$$$\blacktriangle_{{old}} =\mathrm{6}\sqrt{\frac{\mathrm{3}{a}}{\mathrm{2}}\left(\frac{\mathrm{3}{a}}{\mathrm{2}}−{a}\right)^{\mathrm{3}} } \\ $$$$\blacktriangle_{{old}} =\mathrm{6}\sqrt{\frac{\mathrm{3}{a}}{\mathrm{2}}\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{3}} }\:=\mathrm{6}\sqrt{\mathrm{3}\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{4}}…
Question Number 157995 by zainaltanjung last updated on 30/Oct/21 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{following}\:\mathrm{tetra}− \\ $$$$\mathrm{hedral}\:\mathrm{figure}.\:\mathrm{The}\:\mathrm{length}\:\mathrm{of}\: \\ $$$$\mathrm{the}\:\mathrm{rib}\:\mathrm{is}\:\mathrm{10}\:\mathrm{cm}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area} \\ $$$$\:\mathrm{of}\:\:\mathrm{triangle}\:\mathrm{DEF}. \\ $$$$ \\ $$$$ \\ $$ Terms of Service…
Question Number 157991 by zainaltanjung last updated on 30/Oct/21 $$ \\ $$$$ \\ $$$$\mathrm{Length}\:\mathrm{side}\:\mathrm{of}\:\mathrm{hexagonal}\:\mathrm{above}\:\mathrm{is}\:=\mathrm{12}\: \\ $$$$\mathrm{cm}\:.\:\mathrm{Find}\:\mathrm{it}'\mathrm{s}\:\mathrm{total}\:\mathrm{area} \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 157981 by zainaltanjung last updated on 30/Oct/21 $$\mathrm{How}\:\mathrm{much}\:\mathrm{the}\:\mathrm{long}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diagonal} \\ $$$$\:\mathrm{space}\:,\:\mathrm{if}\:\:\mathrm{total}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{is} \\ $$$$\:\mathrm{216}\:\mathrm{cm}^{\mathrm{2}} . \\ $$ Answered by cherokeesay last updated on 30/Oct/21 $${the}\:{area}\:{of}\:{one}\:{face}\:{of}\:{cube}\::…
Question Number 157984 by cherokeesay last updated on 30/Oct/21 Answered by JDamian last updated on 03/Nov/21 $${Assuming}\:\angle{OAB}\:=\:\angle{ABC}\:=\:\mathrm{90}° \\ $$$${there}\:{is}\:{not}\:{such}\:{solution} \\ $$ Commented by ajfour last…
Question Number 157926 by mr W last updated on 29/Oct/21 $${if}\:{the}\:{line}\:{px}+{qy}={r}\:{tangents}\:{the} \\ $$$${ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1},\:{then}\: \\ $$$$\left.\mathrm{1}\right)\:{prove}\:\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{q}}^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \: \\…
Question Number 26693 by Tinkutara last updated on 28/Dec/17 $${STATEMENT}-\mathrm{1}:\:{The}\:{angle}\:{between} \\ $$$${one}\:{of}\:{the}\:{lines}\:{represented}\:{by}\:{ax}^{\mathrm{2}} \:+ \\ $$$$\mathrm{2}{hxy}\:+\:{by}^{\mathrm{2}} \:=\:\mathrm{0}\:{and}\:{one}\:{of}\:{the}\:{lines} \\ $$$${represented}\:{by}\:\left({a}\:+\:\mathrm{2008}\right){x}^{\mathrm{2}} \:+\:\mathrm{2}{hxy} \\ $$$$+\:\left({b}\:+\:\mathrm{2008}\right){y}^{\mathrm{2}} \:=\:\mathrm{0}\:{is}\:{equal}\:{to}\:{angle} \\ $$$${between}\:{other}\:{two}\:{lines}\:{of}\:{the} \\…
Question Number 26449 by Tinkutara last updated on 25/Dec/17 $${Transform}\:{the}\:{equation}\:\mathrm{5}{x}^{\mathrm{2}} \:+\:\mathrm{4}{xy} \\ $$$$+\:\mathrm{2}{y}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{4}{y}\:+\:\mathrm{4}\:=\:\mathrm{0}\:{into}\:{one} \\ $$$${without}\:{xy},\:{x}\:{and}\:{y}\:{terms}. \\ $$ Answered by jota@ last updated on 26/Dec/17…