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Question Number 211122    Answers: 1   Comments: 0

Is there a special rule for divisibility of four-digit numbers by 13?

$$ \\ $$Is there a special rule for divisibility of four-digit numbers by 13?

Question Number 211118    Answers: 0   Comments: 3

Question Number 211115    Answers: 1   Comments: 0

Question Number 211105    Answers: 2   Comments: 0

Question Number 211104    Answers: 0   Comments: 0

Question Number 211099    Answers: 3   Comments: 0

Question Number 211103    Answers: 0   Comments: 0

Question Number 211079    Answers: 1   Comments: 0

Question Number 211075    Answers: 2   Comments: 0

sin (θ+φ)=θ sin θ=φ find θ and φ.

$$\mathrm{sin}\:\left(\theta+\phi\right)=\theta \\ $$$$\mathrm{sin}\:\theta=\phi \\ $$$${find}\:\theta\:{and}\:\phi. \\ $$

Question Number 211058    Answers: 1   Comments: 1

Question Number 211049    Answers: 0   Comments: 0

someone please try question 204666

$$\mathrm{someone}\:\mathrm{please}\:\mathrm{try}\:\mathrm{question}\:\mathrm{204666} \\ $$

Question Number 211043    Answers: 2   Comments: 0

Question Number 211029    Answers: 2   Comments: 0

Question Number 211020    Answers: 1   Comments: 0

Question Number 211019    Answers: 1   Comments: 0

Question Number 211018    Answers: 1   Comments: 1

Question Number 211008    Answers: 0   Comments: 4

Question Number 211006    Answers: 1   Comments: 0

Question Number 210996    Answers: 2   Comments: 0

Question Number 211016    Answers: 3   Comments: 1

Question Number 211004    Answers: 0   Comments: 0

Question Number 211001    Answers: 0   Comments: 5

Question Number 210989    Answers: 1   Comments: 0

prove tan(72^° )=tan(66^° )+tan(36^° )+tan(6^° )

$${prove}\:{tan}\left(\mathrm{72}^{°} \right)={tan}\left(\mathrm{66}^{°} \right)+{tan}\left(\mathrm{36}^{°} \right)+{tan}\left(\mathrm{6}^{°} \right)\:\: \\ $$

Question Number 210987    Answers: 1   Comments: 0

Question Number 210974    Answers: 2   Comments: 0

(√(25))

$$\sqrt{\mathrm{25}} \\ $$

Question Number 210967    Answers: 0   Comments: 0

Q.210956 im read leithold book again , in this book : 1}define : ln(x)=∫_1 ^( x) dx/x x>0 2}define : ln(e)=1=∫_1 ^( e) dx/x 3}define : exp(x)=y ⇔ ln(y)=x ((d(ln(u)))/du)=(1/u) ⇒ ((d(ln(u)))/dx)=((du/dx)/u) ⇒ u=x^r ⇒ ((d(ln(x^r )))/dx)=((rx^(r−1) )/x^r )=r×(1/x)=r×((d(ln(x)))/dx) ⇒ ln(x^r )=rln(x)+K ⇒ x=1 ⇒ K=0 ⇒ ln(x^r )=r×ln(x) ∀x>0 , ∀r get x=e ⇒ ln(e^r )=r×ln(e)=r ⇒ exp(r)=e^r ⇒ exp(x)=e^x =y ∀x log_e (e^x )=log_e (y)=x and define: ln(y)=x ⇒⇒⇒ln(y)=log_e (y)=x

$${Q}.\mathrm{210956} \\ $$$${im}\:{read}\:{leithold}\:{book}\:{again}\:,\:{in}\:{this}\:{book}\:: \\ $$$$\left.\mathrm{1}\right\}{define}\::\:{ln}\left({x}\right)=\int_{\mathrm{1}} ^{\:{x}} {dx}/{x}\:\:\:\:\:\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right\}{define}\::\:{ln}\left({e}\right)=\mathrm{1}=\int_{\mathrm{1}} ^{\:{e}} {dx}/{x} \\ $$$$\left.\mathrm{3}\right\}{define}\::\:{exp}\left({x}\right)={y}\:\Leftrightarrow\:{ln}\left({y}\right)={x} \\ $$$$\frac{{d}\left({ln}\left({u}\right)\right)}{{du}}=\frac{\mathrm{1}}{{u}}\:\Rightarrow\:\frac{{d}\left({ln}\left({u}\right)\right)}{{dx}}=\frac{{du}/{dx}}{{u}}\:\Rightarrow \\ $$$${u}={x}^{{r}} \:\Rightarrow\:\frac{{d}\left({ln}\left({x}^{{r}} \right)\right)}{{dx}}=\frac{{rx}^{{r}−\mathrm{1}} }{{x}^{{r}} }={r}×\frac{\mathrm{1}}{{x}}={r}×\frac{{d}\left({ln}\left({x}\right)\right)}{{dx}} \\ $$$$\Rightarrow\:{ln}\left({x}^{{r}} \right)={rln}\left({x}\right)+{K}\:\Rightarrow\:{x}=\mathrm{1}\:\Rightarrow\:{K}=\mathrm{0} \\ $$$$\Rightarrow\:{ln}\left({x}^{{r}} \right)={r}×{ln}\left({x}\right)\:\:\:\:\:\:\forall{x}>\mathrm{0}\:,\:\forall{r} \\ $$$${get}\:\:{x}={e}\:\:\Rightarrow\:{ln}\left({e}^{{r}} \right)={r}×{ln}\left({e}\right)={r}\:\Rightarrow\:{exp}\left({r}\right)={e}^{{r}} \\ $$$$\Rightarrow\:{exp}\left({x}\right)={e}^{{x}} ={y}\:\:\:\:\:\forall{x} \\ $$$${log}_{{e}} \left({e}^{{x}} \right)={log}_{{e}} \left({y}\right)={x}\:\:\:{and}\:\:\:{define}:\:{ln}\left({y}\right)={x} \\ $$$$\Rightarrow\Rightarrow\Rightarrow{ln}\left({y}\right)={log}_{{e}} \left({y}\right)={x} \\ $$

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