Question Number 203218 by hardmath last updated on 12/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203219 by hardmath last updated on 12/Jan/24 Answered by witcher3 last updated on 15/Jan/24 $$\mathrm{evident} \\ $$$$\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\right)^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{cd}+\mathrm{da}\right) \\…
Question Number 203187 by Amidip last updated on 12/Jan/24 Answered by mr W last updated on 12/Jan/24 $${supposed}:\:{AB}\bot{AC},\:{AD}\bot{BC} \\ $$$$\frac{\mathrm{45}}{{x}}=\frac{{x}+\mathrm{48}}{\mathrm{45}}\: \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{48}{x}−\mathrm{45}^{\mathrm{2}} =\mathrm{0} \\…
Question Number 203208 by hardmath last updated on 12/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203211 by ajfour last updated on 12/Jan/24 Answered by MM42 last updated on 12/Jan/24 $${p}^{\mathrm{2}} =\mathrm{2}−\mathrm{2}{cosa} \\ $$$${q}^{\mathrm{2}} =\mathrm{2}−\mathrm{2}{sina} \\ $$$$\Rightarrow\left({p}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} +\left({q}^{\mathrm{2}}…
Question Number 203206 by hardmath last updated on 12/Jan/24 Answered by mr W last updated on 12/Jan/24 $$\underset{{k}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\mid{i}−{k}\mid \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{i}} {\sum}}\mid{i}−{k}\mid+\underset{{k}={i}+\mathrm{1}} {\overset{\mathrm{100}}…
Question Number 203207 by hardmath last updated on 12/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203202 by Abdullahrussell last updated on 12/Jan/24 Answered by cortano12 last updated on 12/Jan/24 $$\:\:\:\underbrace{\leq} \\ $$ Answered by mr W last updated…
Question Number 203166 by sulaymonnorboyev140 last updated on 11/Jan/24 $$ \\ $$ Answered by MathematicalUser2357 last updated on 15/Jan/24 $$\mathrm{0} \\ $$ Terms of Service…
Question Number 203157 by navin12345 last updated on 11/Jan/24 Answered by witcher3 last updated on 11/Jan/24 $$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} −\mathrm{2x}−\mathrm{2z}+\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}…