Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 145701 by alcohol last updated on 07/Jul/21

Commented by MJS_new last updated on 07/Jul/21

$$1.$$$$\mathrm{square}-2\times (\mathrm{square}-\mathrm{quarter}\mathrm{circle})=$$$$=-\mathrm{square}+2\times \mathrm{quarter}\mathrm{circle}=$$$$=\mathrm{half}\mathrm{circle}-\mathrm{square}=4^2\pi /2-4^2=$$$$=8\pi -16$$$$2.$$$$\mathrm{trapezoid}-\mathrm{rectangular}-\mathrm{half}\mathrm{circle}=$$$$=\frac{6+8}{2}\times 4-4\times 2-2^2\pi /2=20-2\pi$$

Answered by iloveisrael last updated on 07/Jul/21

$$\left(1\right)shadedarea=2(\frac{1}{4}\pi {\left(4\right)}^2-\frac{1}{2}{\left(4\right)}^2)$$$$=2(4\pi -8)=8\pi -16$$

Answered by iloveisrael last updated on 07/Jul/21

$$\left(2\right)shadedarea=\left(\frac{4+6}{2}\right).4-\frac{1}{2}\pi {\left(2\right)}^2$$$$=20-2\pi$$

Commented by alcohol last updated on 07/Jul/21

$$pleaseexplainthis$$$$thanks$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com