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Question-183135

Question Number 183135 by Tawa11 last updated on 21/Dec/22 Answered by TheSupreme last updated on 21/Dec/22 $$\frac{\Delta{y}}{\Delta{h}}=\frac{{y}\left({x}+{h}\right)−{y}\left({x}\right)}{{h}}=\frac{\mathrm{2}\left({x}+{h}\right)^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{h}}{{h}}= \\ $$$$=\frac{\mathrm{2}{h}\left(\mathrm{2}{x}+{h}\right)+\mathrm{5}{h}}{{h}}=\mathrm{4}{x}+\mathrm{5}+\mathrm{2}{h} \\ $$$${x}=\mathrm{2} \\ $$$$\frac{\Delta{y}}{\Delta{h}}=\mathrm{13}+\mathrm{2}{h}\:…

Question-183117

Question Number 183117 by Subhabrata last updated on 20/Dec/22 Answered by Rasheed.Sindhi last updated on 21/Dec/22 $$\begin{array}{|c|c|c|c|c|c|c|c|c|}{{Class}}&\hline{{x}_{{i}} }&\hline{{f}_{{i}} }&\hline{{x}_{{i}} {f}_{{i}} }\\{\mathrm{30}−\mathrm{39}}&\hline{\mathrm{34}.\mathrm{5}}&\hline{\mathrm{2}}&\hline{\mathrm{69}}\\{\mathrm{40}−\mathrm{49}}&\hline{\mathrm{44}.\mathrm{5}}&\hline{\mathrm{3}}&\hline{\mathrm{133}.\mathrm{5}}\\{\mathrm{50}−\mathrm{59}}&\hline{\mathrm{54}.\mathrm{5}}&\hline{\mathrm{11}}&\hline{\mathrm{599}.\mathrm{5}}\\{\mathrm{60}−\mathrm{69}}&\hline{\mathrm{64}.\mathrm{5}}&\hline{\mathrm{20}}&\hline{\mathrm{1290}}\\{\mathrm{70}−\mathrm{79}}&\hline{\mathrm{74}.\mathrm{5}}&\hline{{f}_{\mathrm{5}} }&\hline{\mathrm{74}.\mathrm{5}{f}_{\mathrm{5}} }\\{\mathrm{80}−\mathrm{89}}&\hline{\mathrm{84}.\mathrm{5}}&\hline{\mathrm{25}}&\hline{\mathrm{2112}.\mathrm{5}}\\{\mathrm{90}−\mathrm{99}}&\hline{\mathrm{94}.\mathrm{5}}&\hline{\mathrm{7}}&\hline{\mathrm{661}.\mathrm{5}}\\{}&\hline{}&\hline{\underset{=\mathrm{68}+{f}_{\mathrm{5}} } {\Sigma{f}_{{i}}…

Question-52025

Question Number 52025 by Tawa1 last updated on 02/Jan/19 Commented by Abdo msup. last updated on 03/Jan/19 $$\left.\mathrm{8}\right)\left({x}+\mathrm{1}\right)^{\mathrm{6}} +\left({x}−\mathrm{1}\right)^{\mathrm{6}} =\mathrm{0}\:\Leftrightarrow\frac{\left({x}−\mathrm{1}\right)^{\mathrm{6}} }{\left({x}+\mathrm{1}\right)^{\mathrm{6}} }\:=−\mathrm{1}\:\Leftrightarrow \\ $$$$\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{6}} \:=−\mathrm{1}\:\:{let}\:\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\:={z}\:\Rightarrow{z}^{\mathrm{6}}…