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 62519 by azizullah last updated on 22/Jun/19

Answered by $@ty@m last updated on 22/Jun/19

(i) A=bh−(1/2)bh=(1/2)bh  ⇒2A=bh  (ii) A=ab−4x^2

$$\left({i}\right)\:{A}={bh}−\frac{\mathrm{1}}{\mathrm{2}}{bh}=\frac{\mathrm{1}}{\mathrm{2}}{bh} \\ $$$$\Rightarrow\mathrm{2}{A}={bh} \\ $$$$\left({ii}\right)\:{A}={ab}−\mathrm{4}{x}^{\mathrm{2}} \\ $$

Commented by azizullah last updated on 22/Jun/19

thanks alot

$$\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{alot}}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com