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Question Number 121111 by zakirullah last updated on 05/Nov/20

Commented by benjo_mathlover last updated on 05/Nov/20

$$\left(\mathrm{Q}4\right)\mathrm{n}(\mathrm{A}\cup \mathrm{B})=50+20-10=60$$

Commented by zakirullah last updated on 05/Nov/20

$$greatwork$$

Commented by som(math1967) last updated on 05/Nov/20

$$5.\mathrm{Apple}\mathrm{juice}=\mathrm{A}$$$$\mathrm{Orange}\mathrm{juice}=\mathrm{B}$$$$\mathrm{Taking}\mathrm{neither}\mathrm{Apple}\mathrm{juice}$$$$\mathrm{nor}\mathrm{Orange}\mathrm{juice}=400-\mathrm{n}(\mathrm{A}\cup \mathrm{B})$$$$=400-\{100+150-75\}$$$$=400-175=225$$

Commented by zakirullah last updated on 05/Nov/20

$$\mathit{e}\mathit{x}\mathit{c}\mathit{e}\mathit{l}\mathit{l}\mathit{e}\mathit{n}\mathit{t}\mathit{s}\mathit{i}\mathit{r}!$$

Commented by som(math1967) last updated on 05/Nov/20

$$6.\mathrm{Indian}\mathrm{food}=\mathrm{I}$$$$\mathrm{Fast}\mathrm{food}=\mathrm{F}$$$$\mathrm{n}(\mathrm{I}\cup \mathrm{F})=65$$$$\mathrm{n}\left(\mathrm{I}\right)=40\mathrm{n}(\mathrm{I}\cap \mathrm{F})=10$$$$\mathrm{n}\left(\mathrm{F}\right)=?$$$$65=40+\mathrm{n}\left(\mathrm{F}\right)-10$$$$\therefore \mathrm{n}\left(\mathrm{F}\right)=65-30=35$$$$\therefore \mathrm{only}\mathrm{fast}\mathrm{food}=35-10=25$$

Commented by zakirullah last updated on 05/Nov/20

$$goodwork$$

Commented by zakirullah last updated on 05/Nov/20

$$sircanyousolve?$$$$\mathit{b}\mathit{y}\mathit{r}\mathit{o}\mathit{o}\mathit{t}\mathit{m}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}.$$$$i.f\left(n\right)=6f(n-1)-9f(n-2),f\left(0\right)=1,f\left(1\right)=2$$$$ii.f\left(n\right)=f(n-1)+(n-1),f\left(2\right)=1$$$$iii.f\left(n\right)=3f(n-1),f\left(0\right)=2$$$$iv.f\left(n\right)=3f(n-1)-6f(n-2),f\left(0\right)=2,f\left(1\right)=3$$$$$$

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