Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 129568 by abdurehime last updated on 16/Jan/21

Commented by greg_ed last updated on 16/Jan/21

$$\mathbf{please},\mathbf{review}\mathbf{the}\mathbf{question}2!$$$$\mathbf{it}\mathbf{seems}+\mathbf{or}-\mathbf{is}\mathbf{missed}\mathbf{before}14\mathit{x}.$$

Commented by bemath last updated on 17/Jan/21

$$\left(2\right)\mathcal{L}\equiv {5\mathrm{x}}^2+{5\mathrm{y}}^2+14\mathrm{x}+12\mathrm{y}-10=0$$$$\mathrm{center}\mathrm{point}\{\begin{array}{l}\frac{\partial \mathcal{L}}{\partial \mathrm{x}}=0;10\mathrm{x}+14=0;\mathrm{x}=-\frac{7}{5}\\ \frac{\partial \mathcal{L}}{\partial \mathrm{y}}=0;10\mathrm{y}+12=0;\mathrm{y}=-\frac{6}{5}\end{array}$$$$$$

Commented by abdurehime last updated on 17/Jan/21

$$\mathrm{that}\mathrm{is}+$$

Commented by abdurehime last updated on 17/Jan/21

$$\mathrm{woooooowwww}\mathrm{tnks}$$

Answered by liberty last updated on 16/Jan/21

$$\left(1\mathrm{A}\right)-\frac{5}{\mathrm{B}}=\frac{4}{7};\mathrm{B}=-\frac{35}{4}$$$$\left(1\mathrm{B}\right)(-\frac{5}{\mathrm{B}})\times \left(\frac{4}{7}\right)=-1$$$$\mathrm{B}=\frac{20}{7}$$

Answered by Dwaipayan Shikari last updated on 16/Jan/21

$$5x^2+5y^2\pm 14x+12y-10=0$$$${(x\pm \frac{7}{5})}^2+{(y+\frac{6}{5})}^2=2+\frac{7^2}{5^2}+\frac{6^2}{5^2}$$$$Radius=\frac{135}{25}=5.4$$$$centre=(\pm \frac{7}{5},-\frac{6}{5})$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com