heureka

$$\small{
\begin{array}{llrl}
$(A)$ &\text{Find a number }$0\leq x< &14 &$that solves the congruence $9x \equiv 9 \pmod{14}$.\\
$(B)$&\text{Find a number }$0\leq x< &5 &$that solves the congruence $3x \equiv 4 \pmod{5}$.\\
$(C)$&\text{Find a number }$0\leq x< &1000&$that solves the congruence $999x \equiv 998 \pmod{1000}$.\\
$(D)$&\text{Find a number }$0\leq x< &21 &$that solves the congruence $4x \equiv 17 \pmod{21}$.
\end{array}
}$$

$$\left\{
\begin{array}{l}
$(A)$\\
$(B)$\\
$(C)$\\
$(D)$
\end{array}
\right\}=
\left\{
\begin{array}{lcr}
x&=&1\\
x&=&3\\
x&=&2\\
x&=&20
\end{array}
\right\}, \small{\text{ because }
\left\{
\begin{array}{lcr}
9\cdot 1 &\equiv& 9 \pmod{14}\\
3\cdot 3 &\equiv& 4 \pmod{5}\\
999\cdot 2 &\equiv& 998 \pmod{1000}\\
4\cdot 20 &\equiv& 17 \pmod{21}\\
\end{array}
\right\}$$

(A) Convince yourself that 50 is the inverse of 5 modulo 83, and that 12 is the inverse of 7 modulo 83. Find the inverse of 35 modulo 83. (Give your answer as a nonnegative integer that is less than 83.)

$$\small{\text{
The inverse of $35 \pmod{83} = \mathbf{ 19 }$, because $35 \cdot 19 \equiv 1 \pmod {83}$
}}$$

(b) Find the inverse of 49 modulo 83. (Give your answer as a nonnegative integer that is less than 83.)

$$\small{\text{
The inverse of $49 \pmod{83} = \mathbf{ 61 }$, because $49 \cdot 61 \equiv 1 \pmod {83}$
}}$$

What is the inverse of 100 modulo 10001? (Give your answer as a nonnegative integer that is less than 10001.)

$$\small{\text{
The inverse of $100 \pmod{10001} = \mathbf{ 9901 }$, because $100 \cdot 9901 \equiv 1 \pmod {10001}$
}}$$

1.5=1/cos(x) find x

$$\small{\text{$
\begin{array}{rcl}
\dfrac{ 1 } { \cos{(x)} } &=& 1.5 \\\\
\dfrac{ 1 } { \cos{(x)} } &=& \dfrac{3}{2} \\\\
\cos{(x)} &=& \dfrac{2}{3} \\\\
\cos{(x)} &=& \dfrac{2}{3} \qquad | \qquad \pm\arccos{()}\\\\
x &=& \pm \arccos{ \left(\dfrac{2}{3} \right) }\\\\
x_{1,2} &=& \pm 48.1896851042\ensurement{^{\circ}}\\\\
\mathbf{x_1} & \mathbf{=} & \mathbf{48.1896851042\ensurement{^{\circ}} } \\\\
x_2 &=& -48.1896851042\ensurement{^{\circ}} \\\\
x_2 &=& -48.1896851042\ensurement{^{\circ}} + 360 \ensurement{^{\circ}} \\\\
\mathbf{x_2} & \mathbf{=} & \mathbf{311.810314896\ensurement{^{\circ}} }
\end{array}
$}}$$

You can get 28200 if u dont reduce it with 5 % the last day.

You get max 20 * 1410 = 28200.

$$\small{\text{$
\boxed{~~t_n=0.95^n\cdot \left( \dfrac{25547}{0.95}-20\cdot 1410 \right)+20\cdot 1410~~}
$}}\\\\
\small{\text{We have $t_n = 27924$ and get n by rearrange:
}}\\\\
\small{\text{$
\boxed{
n=\dfrac
{\ln{\left( \dfrac{t_n -20\cdot 1410}{ \dfrac{25547}{0.95} -20\cdot 1410} \right)}}
{\ln{(0.95)}}
}
$}}\\\\\\
\small{\text{$
\begin{array}{rcl}
n &=& \dfrac
{\ln{\left( \dfrac{27924 -20\cdot 1410}{\dfrac{25547}{0.95}-20\cdot 1410} \right)}}
{\ln{(0.95)}}\\\\
n &=& \dfrac
{\ln{\left( \dfrac{-276}{ -1308.42105263} \right)}}
{\ln{(0.95)}}\\\\
n &=& \dfrac
{\ln{\left( 0.21094127112\right)}}
{\ln{(0.95)}}\\\\
n &=& \dfrac
{-1.55617552018}
{-0.05129329439}\\\\
\mathbf{n} & \mathbf{=} & \mathbf{30.3387711544}
\end{array}
$}}$$

It will take to get 27924 about 30.3387711544 days

day 30 = 27919.16 

day 31 = 27933.20


$$\small{\text{$ 
 \begin{array}{lr}
 \small{\text{day }} 1 & 26957.00 \\
 \small{\text{day }} 2 & 27019.15 \\
 \small{\text{day }} 3 & 27078.19 \\
 \small{\text{day }} 4 & 27134.28 \\
 \small{\text{day }} 5 & 27187.57 \\
 \small{\text{day }} 6 & 27238.19 \\
 \small{\text{day }} 7 & 27286.28 \\
 \small{\text{day }} 8 & 27331.97 \\ 
 \small{\text{day }} 9 & 27375.37 \\
 \small{\text{day }} 10 & 27416.60 \\
 \small{\text{day }} 11 & 27455.77 \\
 \small{\text{day }} 12 & 27492.98 \\
 \small{\text{day }} 13 & 27528.33 \\
 \small{\text{day }} 14 & 27561.92 \\
 \small{\text{day }} 15 & 27593.82 \\
 \small{\text{day }} 16 & 27624.13 \\
 \small{\text{day }} 17 & 27652.92 \\
 \small{\text{day }} 18 & 27680.28 \\
 \small{\text{day }} 19 & 27706.26 \\
 \small{\text{day }} 20 & 27730.95 \\
 \small{\text{day }} 21 & 27754.40 \\
 \small{\text{day }} 22 & 27776.68 \\
 \small{\text{day }} 23 & 27797.85 \\
 \small{\text{day }} 24 & 27817.96 \\
 \small{\text{day }} 25 & 27837.06 \\
 \small{\text{day }} 26 & 27855.20 \\
 \small{\text{day }} 27 & 27872.44 \\
 \small{\text{day }} 28 & 27888.82 \\
 \small{\text{day }} 29 & 27904.38 \\
 \small{\text{day }} 30 & 27919.16 \\
 \small{\text{day }} 31 & 27933.20
 \end{array}
 $}}$$

if i have 25547 and add 1410 to that sum i get = 26957. if i reduce that new sum with 5 % i get = 25609,15. if i repeat this process each day, how many days will it take to get to 27924?

If I have properly understood: You get max 19 * 1410 = 26790

(19 = 0.95/0.05)

You never reach 27924!

$$\small{\text{$
\begin{array}{rcl}
\dfrac{ 3^{4n}-3^{2n} } { 3^{3n} + 3^{2n} } \\
&=& \dfrac{ 3^{2n+2n}-3^{2n} } { 3^{2n+n} + 3^{2n} } \\\\
&=& \dfrac{ 3^{2n}3^{2n}-3^{2n} } { 3^{2n}3^{n} + 3^{2n} } \\\\
&=& \dfrac{ 3^{2n} (3^{2n}-1) } { 3^{2n} (3^{n} + 1) } \\\\
&=& \dfrac{ 3^{2n}-1 } { 3^{n}+1 } \\\\
&=& \dfrac{ (3^{n}-1)(3^{n}+1) } { 3^{n}+1 } \\\\
\mathbf{ \dfrac{ 3^{4n}-3^{2n} } { 3^{3n} + 3^{2n} } }&\mathbf{=} & \mathbf{3^{n}-1}
\end{array}
$}}$$

$$3.\\\\
\small{\text{$
\dfrac{ \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)
*
\left( \dfrac{ x^2-xy }{ x^4-y^4 } \right) }
{ \dfrac{ x } { x^2+2xy+y^2 } }
$}}$$

$$\small{\text{$
\begin{array}{rcl}
&=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)*
\left( \dfrac{ x^2-xy }{ x^4-y^4 } \right) *
\left( \dfrac { x^2+2xy+y^2 } { x } \right) \\\\
&=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)*
\left( \dfrac{ x(x-y) }{ x^4-y^4 } \right) *
\left( \dfrac { (x+y)^2 } { x } \right) \\\\
&=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)*
\left( \dfrac{ (x-y)(x+y)^2 }{ x^4-y^4 } \right)\\\\
&=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)*
\left( \dfrac{ (x-y)(x+y)(x+y) }{ x^4-y^4 } \right)\\\\
&=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)*
\left( \dfrac{ (x^2-y^2)(x+y) }{ x^4-y^4 } \right)\\\\
&=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)*
\left( \dfrac{ (x^2-y^2)(x+y) }{ (x^2-y^2)(x^2+y^2) } \right)\\\\
&=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)*
\left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\
&=& \left( \dfrac{x(x-y)+y(x+y)}{(x-y)(x+y)}\right)*
\left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\
&=& \left( \dfrac{ x^2+y^2 }{(x-y)(x+y)}\right)*
\left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\
&\mathbf{=} & \mathbf{ \dfrac{ 1 }{x-y} }
\end{array}
$}}$$

Find The age for a rock for which you determine that 65% as the original uranium – 238 remainswhile the other 35% has decayed into lead

I assume: uranium-lead decay route $${}^{238}U$$ to $${}^{206}{Pb}$$

(i)

$$\boxed{~N_t = N_0\cdot e^{-k\cdot t} ~~(1)~} \quad \small{\text{k, is called the decay constant.}}\\ \small{\text{$N_0$ is the initial number of nuclei (at time zero), and}} \\ \small{\text{$N_t$ is the number remaining after the time interval.}} \\ \small{\text{We can also write our equation:~}}
\boxed{~\ln{ \left( \dfrac{N_t}{N_0} } \right) = -k\cdot t ~~(2)~}$$

(ii)

$$\small{\text{Relationship between the decay constant, k, and half-life, $t_{1/2}$.}}\\
\small{\text{We set $N_{t_{1/2}} = \dfrac{N_0}{2} $ in (1)~so we have $ \dfrac{N_0}{2}=N_0\cdot e^{-k\cdot t_{1/2}} $ and get }}\\
\boxed{~ k = \dfrac{ \ln{(2)} }{ t_{1/2} } ~~(3)~}$$

(iii)

$$\small{\text{We get t, if we rearrange (2). We have~~}}
\boxed{~ t = -\dfrac{1}{k} \cdot\ln{ \left( \dfrac{N_t}{N_0} } \right)~~(4)~}\\
\small{\text{We set k in (3) into (4). We have~~}}
\boxed{~ t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{N_t}{N_0} } \right)~~(5)~}\\$$

(iv)

$$\small{\text{$N_0 = {}^{238}U + {}^{206}Pb\cdot \dfrac{238}{206} $. And we have $35\%$ of ${}^{238}U$ is ${}^{206}Pb$ so }}\\
\small{\text{$N_0 = {}^{238}U + ({}^{238}U\cdot 35\%) \cdot \dfrac{238}{206}. $We obtain the latter quantity of ${}^{206}Pb$ by}}\\
\small{\text{multiplying the present mass of lead-206 by the ratio of the atomic mass of uranium}}\\
\small{\text{to that of lead, into which it has decayed. $N_t ={}^{238}U$. }}$$

$$\small{\text{ Finally
$\dfrac{N_t}{N_0} = \dfrac{ {}^{238}U } {{}^{238}U + ({}^{238}U\cdot 35\%) \cdot \dfrac{238}{206}} = \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} }
$
}}\\
\boxed{ \dfrac{N_t}{N_0} = \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} }
}$$

(v)

$$\small{\text{If we regard the half-life $t_{1/2}$ for the decay of}}\\
\small{\text{uranium-238 $({}^{238}U)$ to lead-206 $({}^{206}Pb)$ is $\mathbf{4.468\cdot 10^9~ yr }$, }}\\
\small{\text{we can calculate how old the rock is.}}\\
\small{\text{We now have
$t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} } } \right)
$
}}\\
\small{\text{or $t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} } } \right)
$
}} \\
}}\\
\small{\text{or $t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot
\left[ \ln{ (1) } -
\ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right)
\right]
$
}} \\
}}\\
\small{\text{Because $\ln{(1)}= 0$
}} \\
\small{\text{$t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot
\left[ 0 -
\ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right)
\right]
$
}} \\\\
\small{\text{and finally}}\\\\
\boxed{t = t_{1/2} \cdot \dfrac
{ \ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right) }
{ \ln(2) }}$$

(vi)

$$\small{\text{We calculate:}}\\
\small{\text{$
\begin{array}{rcl}
t &=&
t_{1/2} \cdot
\dfrac
{ \ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right) }
{ \ln(2) }\\
t &=& 4.468\cdot 10^9~ yr \cdot
\dfrac
{ \ln{ \left( 1.40436893204} \right) }
{ \ln(2) }\\\\
t &=& 4.468\cdot 10^9~ yr \cdot
\dfrac
{ 0.33958804319 }
{0.69314718056 }\\\\
t &=& 4.468\cdot 10^9~ yr \cdot 0.48992198586 \\\\
\mathbf{t} &\mathbf{=}& \mathbf{ 2.18897143282\cdot 10^9~ yr }
\end{array}
$}}$$

The rock is 2 188 971 432.82 years old 

Hallo sabi,

your way is not wrong.