Question Number 228474 by ajfour last updated on 18/Apr/26 Commented by ajfour last updated on 18/Apr/26 $${Parabola}\:{is}\:\:{y}=\left({h}−{x}\right)^{\mathrm{2}} \\ $$$${Find}\:{circle}\:{radius}\:{in}\:{terms}\:{of}\:{h}. \\ $$$${How}\:{much}\:{is}\:{r}\:{for}\:{h}=\mathrm{1}? \\ $$ Commented by…
Question Number 228378 by ajfour last updated on 10/Apr/26 Commented by ajfour last updated on 10/Apr/26 $${parabola}\:{eq}.\:{is}\:{y}={b}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$${Find}\:{radius}\:{of}\:{circle}\:{in}\:{terms} \\ $$$${of}\:{b}. \\ $$ Answered…
Question Number 228342 by ajfour last updated on 08/Apr/26 Commented by ajfour last updated on 08/Apr/26 $${If}\:\:{for}\:{rectangles}\:{ABCD}\:{and} \\ $$$${PQRS}:\:\:\:\:\:\:{AB}=\mathrm{2}{a},\:{BC}=\mathrm{2}{b}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{PQ}={a},\:{QR}={b} \\ $$$${find}\:{lengths}\:{BE},\:{DF}. \\ $$…
Question Number 227326 by gregori last updated on 16/Jan/26 $$\:\:\:\mid\:{x}+\mathrm{1}\:\mid\:+\:\mid\:{x}\:\mid\:+\:\mid\:{x}−\mathrm{3}\mid\:>\:\mathrm{8}\:\: \\ $$ Answered by Kassista last updated on 16/Jan/26 $$ \\ $$$${if}\:{x}\geqslant\mathrm{3}: \\ $$$${x}+\mathrm{1}+{x}+{x}−\mathrm{3}>\mathrm{8}\:\Leftrightarrow\:\mathrm{3}{x}−\mathrm{2}>\mathrm{8},\:\mathrm{3}{x}>\mathrm{10},\:{x}>\frac{\mathrm{10}}{\mathrm{3}}…\left(\mathrm{1}\right) \\…
Question Number 227294 by Spillover last updated on 12/Jan/26 Commented by Spillover last updated on 12/Jan/26 $${find}\:{shaded}\:{area}\:{of}\:{rectangle}\:{above} \\ $$ Answered by Spillover last updated on…
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Question Number 226829 by Estevao last updated on 16/Dec/25 Answered by Estevao last updated on 17/Dec/25 $${Good} \\ $$ Answered by mr W last updated…
Question Number 226770 by gregori last updated on 14/Dec/25 Answered by TonyCWX last updated on 14/Dec/25 $$\mathrm{Stewart}'\mathrm{s}\:\mathrm{Theorem}: \\ $$$${b}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {y}=\left({x}+{y}\right)\left({z}^{\mathrm{2}} −{xy}\right) \\ $$$${b}^{\mathrm{2}} \left({x}+{y}\right)=\left({x}+{y}\right)\left({z}^{\mathrm{2}}…