heureka

Right triangle ABC has legs of length AB = 12 and BC = 6.

Point D is randomly chosen within the triangle ABC.

What is the probability that the triangle ABD has area less than or equal to 12 ?

$\begin{array}{|rcll|} \hline ABC_{\text{Area } } &=& \frac{AB \cdot BC }{2} \quad & | \quad AB=12 \quad BC = 6 \\ &=& \frac{12 \cdot 6 }{2} \\ &=& 36 \\ \hline \end{array}$

Let d = distance parallel from leg AB.

$\begin{array}{rcll} ABD_{\text{Area } } = \frac{AB \cdot d }{2} &=& 12 \quad & | \quad AB=12 \\ \frac{12 \cdot d }{2} &=& 12 \\ \frac{d }{2} &=& 1 \\ d &=& 2 \\ \end{array}$

$\begin{array}{|rcll|} \hline \frac{BC-d}{x} &=& \frac{BC}{AB} \quad & | \quad AB=12 \quad BC = 6 \quad d = 2 \\ \frac{6-2}{x} &=& \frac{6}{12} \\ \frac{4}{x} &=& \frac{1}{2} \\ x &=& 8 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline ABD_{\text{Area } } = Area_{\text{trapezoid } } &=& (\frac{AB+x}{2}) \cdot d \quad & | \quad AB=12 \quad x = 8 \quad d = 2 \\ &=& (\frac{12+8}{2}) \cdot 2 \\ &=& 20 \\ \hline \end{array}$

probability:

$\begin{array}{|rcll|} \hline \frac{ ABD_{\text{Area } } } {ABC_{\text{Area } }} &=& \frac{20}{36} \\ &=& \frac{5}{9} \quad (55.\bar{5}\ \% ) \\ \hline \end{array}$

Hi - what is the best way to solve simultaneous equations where one is quadratic and the other linear?

For example:

y=x+3

y=x²+3x

1. Formula line ( linear equation)

$y_{\text{line}}=m\cdot x_{\text{line}}+b$

2. Formula parabola (quadatic equation)

$y_{\text{parabola}}=A\cdot x_{\text{parabola}}^2+B\cdot x_{\text{parabola}} + C$

3. set equal  $y_{\text{line}}=y_{\text{parabola}}=y_{\text{intersection}}$ :

$\begin{array}{|rcll|} \hline m\cdot x_{\text{intersection}}+b &=& A\cdot x_{\text{intersection}}^2+B\cdot x_{\text{intersection}} + C \\\\ Ax_{\text{intersection}}^2+x_{\text{intersection}}(B-m)+C-b &=& 0 \\ x_{\text{intersection}_{1,2}} &=& \dfrac{m-B\pm \sqrt{(m-B)^2-4\cdot A \cdot(C-b)} }{2A} \\ y_{\text{intersection}_{1,2}} &=& m\cdot x_{\text{intersection}_{1,2}} + b \\ \hline \end{array}$

4. Example:

$\begin{array}{|rcll|} \hline y &=& x + 3 \quad & \quad m=1 \quad b = 3 \\ y &=& x^2+3x \quad & \quad A=1 \quad B = 3 \quad C = 0 \\ x_{\text{intersection}_{1,2}} &=& \dfrac{1-3\pm \sqrt{(1-3)^2-4\cdot 1 \cdot(0-3)} }{2\cdot 1} \\ x_{\text{intersection}_{1,2}} &=& \dfrac{-2\pm \sqrt{4+12} }{2} \\ x_{\text{intersection}_{1,2}} &=& \dfrac{-2\pm 4 }{2} \\ x_{\text{intersection}_{1}} &=& \dfrac{-2 + 4 }{2} \\ &=& 1 \\\\ x_{\text{intersection}_{2}} &=& \dfrac{-2 - 4 }{2} \\ &=& -3 \\\\ y_{\text{intersection}_{1}} &=& 1\cdot x_{\text{intersection}_{1}} + 3 \\ &=& 1\cdot 1 + 3 \\ &=& 4 \\\\ y_{\text{intersection}_{2}} &=& 1\cdot x_{\text{intersection}_{2}} + 3 \\ &=& 1\cdot (-3) + 3 \\ &=& 0 \\ \hline \end{array}$

winkel aus tangens ermitteln

Mit Hilfe des Arkustangens kann ${\displaystyle \varphi }$ wie folgt  bestimmt werden:

${\displaystyle \varphi ={ \begin{cases} \arctan {(\frac {y}{x})}&\mathrm {f{\ddot {u}}r} \ x>0\\ \arctan {(\frac {y}{x})}+180^{\circ} &\mathrm {f{\ddot {u}}r} \ x<0,\ y\geq 0\\ \arctan {(\frac {y}{x})}-180^{\circ} &\mathrm {f{\ddot {u}}r} \ x<0,\ y<0\\ 90^{\circ} &\mathrm {f{\ddot {u}}r} \ x=0,\ y>0\\ 270^{\circ} &\mathrm {f{\ddot {u}}r} \ x=0,\ y<0\\ \end{cases}}}$

Probability Question

Daniel has a bag of marbles. he has twice as many black marbles as red marbles.

the rest are yellow.

he is going to take a marble at random from the bag.

yellow is 1/7

what is black and red probability?

$\begin{array}{|rcll|} \hline b=2r \\ \hline \end{array}$

$\begin{array}{|lcrcll|} \hline \text{Probability yellow } = \frac17 & \Rightarrow & y+b+r &=& 7 & \text{ and } \quad y = 1 \\ & & y+b+r &=& 7 & \qquad b=2r \quad y = 1 \\ & & 1+2r+r &=& 7 \\ & & 3r &=& 6 \\ & & \mathbf{ r } & \mathbf{=} & \mathbf{2} \\ \text{Probability red:} & & \mathbf{ \frac{r}{y+b+r} } & \mathbf{=} & \mathbf{\frac27 } \\\\ & & y+b+r &=& 7 & \qquad r=2 \quad y = 1 \\ & & 1+b+2 &=& 7 & \\ & & b &=& 7-3 & \\ & & \mathbf{ b } & \mathbf{=} & \mathbf{4} \\ \text{Probability black:} & & \mathbf{ \frac{b}{y+b+r} } & \mathbf{=} & \mathbf{\frac47 } \\\\ \hline \end{array}$

$\begin{array}{|l|r|r|r|} \hline \text{Colour} & \text{yellow} & \text{black} & \text{red} \\ \hline \text{Probability} & \frac17 & \frac47 & \frac27 \\ \hline \end{array}$

Integrate the following with steps please:

integrate [sin^2 (x) cos^2 (x)] dx from x=0 to pi.

Thank you for any help.

Formula:

$\begin{array}{|rcll|} \hline \cos^2(x) = \frac12\cdot \Big(1+\cos(2x)\Big) \\ \sin^2(x) = \frac12\cdot \Big(1-\cos(2x)\Big) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \int \limits_{0}^{\pi} {\sin^2(x) \cos^2(x)\ dx} \\ &=& \int \limits_{0}^{\pi} {\frac12\cdot \Big(1-\cos(2x)\Big)\cdot \frac12\cdot \Big(1+\cos(2x)\Big)\ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\cos(2x)\Big)\cdot \Big(1+\cos(2x)\Big)\ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\cos^2(2x)\Big) \ dx} \quad & | \quad \cos^2(2x) = \frac12\cdot \Big(1+\cos(4x)\Big) \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\frac12\cdot \Big(1+\cos(4x)\Big) \ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big(1-\frac12 - \cos(4x)\Big) \ dx} \\ &=& \frac14\cdot \int \limits_{0}^{\pi} { \Big( \frac12 - \cos(4x)\Big) \ dx} \\ &=& \frac14\cdot [\frac{x}{2}-\frac{\sin(4x)}{4}]_{0}^{\pi} \\ &=& \frac18\cdot [ x - \frac{\sin(4x)}{2} ]_{0}^{\pi} \\ &=& \frac18\cdot [ \pi - \sin(4\pi)-(0-\frac{\sin(4\cdot 0)}{2} ) ] \\ &=& \frac18\cdot [ \pi - 0-(0-0) ] \\ &=& \frac18\cdot [ \pi ] \\ &=& \frac{\pi}{8}\\ \hline \end{array}$

*edited

How many 3s..........

How many digits of 3 are there between 1 and 10^12, and what is the percentage of these 3's out of the total of 10^12 numbers, and why? Please explain.

Thank you for any help.

I agree with Alan the total count of 3s is in general 

between 1 and 10ⁿ =  $n\cdot 10^{n-1}$

between 1 and 10¹²  = $12\cdot 10^{11}$

My empirical solution:

$\small{ \begin{array}{|lcr|c|rrrrrrrrrrrrr|r|} \hline &&& n & &12th&11th&10th&9th&8th&7th&6th&5th&4th&third&second&first & \text{Sum of 3's} \\ &&& & \text{3's in Postion} \\ \hline 1 &\ldots& 10 & 1 & & & & & & & & & & & & &1\times 3 & 1\\ 1 &\ldots& 100 & 2 & & & & & & & & & & & &10\times 3 &10\times 3 & 20 \\ 1 &\ldots& 1000 & 3 & & & & & & & & & & &100\times 3&100\times 3&100\times 3 & 300 \\ 1 &\ldots& 10000 & 4 & & & & & & & & & &1000\times 3&1000\times 3&1000\times 3&1000\times 3 & 4000 \\ \cdots \\ 1 &\ldots& 1000000000000 & 12 & &100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3&100000000000\times 3 & 1200000000000 \\ \hline \end{array} }$

3^(x+1)=2^(x+2)

$\begin{array}{|rcll|} \hline 3^{x+1}& = & 2^{x+2} \\ 3^{x+1}& = & 2^{x+1}\cdot 2^1 \quad & | \quad : 2^{x+1} \\ \frac{ 3^{x+1} } {2^{x+1}} & = & 2 \\ (\frac32)^{x+1} & = & 2 \\ (1.5)^{{x+1}} & = & 2 \quad & | \quad \ln \text{both sides} \\ \ln\Big((1.5)^{x+1} \Big) & = & \ln(2) \quad & | \quad \ln(a^b) = b\cdot \ln(a) \\ (x+1)\cdot \ln(1.5) & = & \ln(2) \quad & | \quad : \ln(1.5) \\ x+1& = & \frac{\ln(2)} { \ln(1.5) } \quad & | \quad -1 \\ x & = & \frac{\ln(2)} { \ln(1.5) } -1 \\ x & = & 0.70951129135 \\ \hline \end{array}$

I want to caculate 11power 850 mod 1643 

$\begin{array}{|rcll|} \hline && 11^{850} \pmod{1643} \\ & \equiv & 11^{4\cdot 212+2} \pmod{1643} \\ & \equiv & (11^4)^{212}\cdot 121 \pmod{1643} \quad & | \quad 11^4\pmod{1643} &\equiv& -146 \pmod{1643} \\ & \equiv & (-146)^{212}\cdot 121 \pmod{1643} \\ & \equiv & (-146)^{2\cdot 106}\cdot 121 \pmod{1643} \\ & \equiv & [(-146)^2]^{106}\cdot 121 \pmod{1643} \quad & | \quad (-146)^2\pmod{1643} &\equiv& -43 \pmod{1643} \\ & \equiv & (-43)^{106}\cdot 121 \pmod{1643} \\ & \equiv & (-43)^{4\cdot 26+2}\cdot 121 \pmod{1643} \\ & \equiv & [(-43)^4]^{26}\cdot (-43)^2 \cdot 121 \pmod{1643} \quad & | \quad (-43)^4 \pmod{1643} &\equiv& -282 \pmod{1643} \\ & \equiv & (-282)^{26}\cdot (-43)^2 \cdot 121 \pmod{1643} \quad & | \quad (-43)^2 \pmod{1643} &\equiv& 206 \pmod{1643} \\ & \equiv & (-282)^{26}\cdot 206 \cdot 121 \pmod{1643} \\ & \equiv & (-282)^{2\cdot 13}\cdot 206 \cdot 121 \pmod{1643} \\ & \equiv & [(-282)^2]^{13} \cdot 206 \cdot 121 \pmod{1643} \quad & | \quad (-282)^2 \pmod{1643} &\equiv& 660 \pmod{1643} \\ & \equiv & 660^{13} \cdot 206 \cdot 121 \pmod{1643} \\ & \equiv & 660^{2\cdot6 + 1} \cdot 206 \cdot 121 \pmod{1643} \\ & \equiv & (660^2)^6\cdot 660 \cdot 206 \cdot 121 \pmod{1643} \quad & | \quad 660^2\pmod{1643} &\equiv& 205 \pmod{1643} \\ & \equiv & 205^6\cdot 660 \cdot 206 \cdot 121 \pmod{1643} \quad & | \quad 660 \cdot 206 \cdot 121\pmod{1643} &\equiv& -199 \pmod{1643} \\ & \equiv & 205^6\cdot (-199) \pmod{1643} \\ & \equiv & 205^{2\cdot 3}\cdot (-199) \pmod{1643} \\ & \equiv & (205^2)^3\cdot (-199) \pmod{1643} \quad & | \quad 205^2\pmod{1643} &\equiv& -693 \pmod{1643} \\ & \equiv & (-693)^3\cdot (-199) \pmod{1643} \\ & \equiv & (-693)^{2+1}\cdot (-199) \pmod{1643} \\ & \equiv & (-693)^2\cdot(-693)\cdot (-199) \pmod{1643} \quad & | \quad (-693)\cdot (-199)\pmod{1643} &\equiv& -105 \pmod{1643} \\ & \equiv & (-693)^2\cdot(-105) \pmod{1643} \quad & | \quad (-693)^2\pmod{1643} &\equiv& 493 \pmod{1643} \\ & \equiv & 493\cdot(-105) \pmod{1643} \\ & \equiv & 493\cdot(-105) \pmod{1643} \quad & | \quad 493\cdot(-105) \pmod{1643} &\equiv& -832 \pmod{1643} &\equiv& 811 \pmod{1643} \\ & \equiv & 811 \pmod{1643} \\ \hline \end{array}$

what is 0.0992125657 as a fraction

$\begin{array}{|rcll|} \hline 0.0992125657 & \approx & \frac{878473}{8854453} \\ & \approx & {\color{red}0.0992125656}99992986579746936372015301227529244324861174371\ldots \\ \hline \end{array}$

e^ln z

$e ^{\ln (z)} = z$