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Author: Tinku Tara

The-acute-angle-of-the-rectangle-trapezius-is-equal-to-90-arcsin0-1-The-bases-measure-10-and-30-Calculate-the-area-of-the-trapezius-

Question Number 72998 by yannickmendes_33 last updated on 05/Nov/19 $${The}\:{acute}\:{angle}\:{of}\:{the}\:{rectangle}\:{trapezius}\:{is}\:{equal}\:{to}\:\alpha=\mathrm{90}°{arcsin}\mathrm{0}.\mathrm{1} \\ $$$${The}\:{bases}\:{measure}\:\mathrm{10}\:{and}\:\mathrm{30}.\:{Calculate}\:{the}\:{area}\:{of}\:{the}\:{trapezius}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

x-y-7-x-y-11-faind-x-and-y-

Question Number 138532 by mathlove last updated on 14/Apr/21 $${x}+\sqrt{{y}}=\mathrm{7} \\ $$$$\sqrt{{x}}+{y}=\mathrm{11} \\ $$$${faind}\:\:{x}=?\:\:{and}\:\:{y}=? \\ $$ Answered by henderson last updated on 14/Apr/21 $$\begin{cases}{{x}\:+\:\sqrt{{y}\:}\:=\:\mathrm{7}}\\{\sqrt{{x}}\:+\:{y}\:=\:\mathrm{11}}\end{cases}\:\Leftrightarrow\:\begin{cases}{{x}\:=\:\mathrm{7}\:−\:\sqrt{{y}}}\\{\sqrt{\mathrm{7}−\sqrt{{y}}}\:+\:{y}\:=\:\mathrm{11}}\end{cases}\:\Leftrightarrow\:\begin{cases}{{x}\:=\:\mathrm{7}−\sqrt{{y}}}\\{\sqrt{\mathrm{7}−\sqrt{{y}}}\:=\:\mathrm{11}−{y}}\end{cases}\Leftrightarrow \\…

The-area-of-the-equilateral-triangle-is-equal-to-16-8-3-pi-Calculate-the-area-of-the-circle-inscribed-in-the-triangle-

Question Number 72997 by yannickmendes_33 last updated on 05/Nov/19 $${The}\:{area}\:{of}\:{the}\:{equilateral}\:{triangle}\:{is}\:{equal}\:{to}\:\frac{\sqrt{\mathrm{16}}\sqrt{\mathrm{8}}}{\mathrm{3}\sqrt{\pi}} \\ $$$${Calculate}\:{the}\:{area}\:{of}\:{the}\:{circle}\:{inscribed}\:{in}\:{the}\:{triangle}. \\ $$$$\: \\ $$ Answered by Kunal12588 last updated on 05/Nov/19 $${area}\:{of}\:{equilateral}\:\bigtriangleup\:=\:\frac{\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{\sqrt{\mathrm{16}}\sqrt{\mathrm{8}}}{\mathrm{3}\sqrt{\pi}}=\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{\pi}}…

advanced-calculus-I-pi-2-pi-2-sin-2-tan-x-dx-pi-e-sinh-1-

Question Number 138524 by mnjuly1970 last updated on 14/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…….{advanced}\:…\:….\:…\:{calculus}….. \\ $$$$\:\:\:\boldsymbol{\mathrm{I}}:=\int_{\frac{−\pi}{\mathrm{2}}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} \left({tan}\left({x}\right)\right){dx}\overset{???} {=}\frac{\pi}{{e}}{sinh}\left(\mathrm{1}\right) \\ $$ Answered by Dwaipayan Shikari last updated on…