Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 44200 by Tawa1 last updated on 23/Sep/18

Commented by MJS last updated on 23/Sep/18

$$A$$$$term1\times term2=0\iff term1=0\vee term2=0$$$${\mathrm{e}}^x-{\mathrm{e}}^{\pi }=0\Rightarrow x=\pi$$$${\mathrm{e}}^x-\pi =0\Rightarrow x=\ln \pi$$$$B$$$$n=2k[\Rightarrow n\mathrm{even}]$$$$n^4-n^3+n^2-n=16k^4-8k^3+4k^2-2k=$$$$=2(8k^4-4k^3+2k^2-k)\mathrm{which}\mathrm{is}\mathrm{divisible}\mathrm{by}2$$$$n=2k+1[\Rightarrow n\mathrm{uneven}]$$$$n^4-n^3+n^2+-n=16k^4+24k^3+16k^2+4k=$$$$=4(4k^4+6k^3+4k^2+k)\mathrm{which}\mathrm{is}\mathrm{divisible}\mathrm{by}2$$$$C$$$$\frac{4\pi }{3}{r}^3={\pi }^3\Rightarrow r=\frac{1}{2}\sqrt[3]{6{\pi }^2}$$$$\mathrm{you}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{build}\mathrm{this}\mathrm{because}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{exact}$$$$\mathrm{value},\mathrm{neither}\mathrm{for}r\mathrm{nor}\mathrm{for}\mathrm{perimeter}$$$$D$$$${\log }_2(2^2+5\times 2^2\times 3)={\log }_2\left(2^2(1+15)\right)=$$$$={\log }_2\left(2^2\times 2^4\right)={\log }_2{2}^6=6$$$${2\log }_{3}2+{\log }_3(7-\frac{1}{4})={2\log }_{3}2+{\log }_3\frac{27}{4}=$$$$={2\log }_{3}2+{\log }_{3}27-{\log }_{3}4=$$$$={2\log }_{3}2+{\log }_3{3}^3-{\log }_3{2}^2=$$$$={2\log }_{3}2+3-{2\log }_{3}2=3$$$$\frac{{({\log }_{2}128-2)}^3}{3+2}=\frac{{({\log }_2{2}^7-2)}^3}{5}=\frac{{(7-2)}^3}{5}=25$$$${(-1)}^{32+{\pi }^0}={(-1)}^{32+1}={(-1)}^{33}=-1$$$$6\times 3+25-1=42$$$$E$$$$\mathrm{area}=\mathrm{rectangle}+\mathrm{triangle}$$$$\mathrm{rectangle}=a\times y$$$$\mathrm{triangle}=\frac{a\times (x-y)}{2}$$$$\mathrm{area}=\frac{a\times (x+y)}{2}$$

Commented by maxmathsup by imad last updated on 23/Sep/18

$$youareasuperherothankyousir...$$

Commented by Tawa1 last updated on 23/Sep/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{really}\mathrm{appreciate}\mathrm{your}\mathrm{effort}.$$

Commented by MJS last updated on 23/Sep/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com