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2)
A boat traveled 60 miles downstream and back. The trip downstream took 3 hours.
The trip back took 6 hours.
Find the speed of the boat in still water and the speed of the current.

x = speed of the boat in still water.

y = speed of the current.

s = 60 miles.

\(\begin{array}{|lrcll|} \hline (1) \text{ downstream} & s &=& (x+y)\cdot 3\ h \\ (2) \text{ back} & s &=& (x-y)\cdot 6\ h \\ \hline & s = (x+y)\cdot 3 &=& (x-y)\cdot 6\\ & (x+y)\cdot 3 &=& (x-y)\cdot 6 \quad & | \quad : 3 \\ & x+y &=& 2(x-y) \\ & x+y &=& 2x-2y \quad & | \quad -2x \\ & -x+y &=& -2y \quad & | \quad -y \\ & -x &=& -3y \\ & \mathbf{x} & \mathbf{=} & \mathbf{3y} \\\\ & s &=& (x+y)\cdot 3\ h \quad & | \quad s = 60\ \text{miles} \\ & 60 &=& (3y+y)\cdot 3 \\ & 60 &=& 4y\cdot 3 \quad & | \quad : 3 \\ & 20 &=& 4y \quad & | \quad : 4 \\ & 5 &=& y \\ & \mathbf{y} & \mathbf{=} & \mathbf{5} \\\\ & x & = & 3y \quad & | \quad y = 5 \\ & x & = & 3\cdot 5 \\ & \mathbf{x} & \mathbf{=} & \mathbf{15} \\ \hline \end{array}\)

The speed of the boat in still water is  \(\mathbf{15\ \dfrac{\text{miles}}{\text{hours}}}\)
The speed of the current is  \(\mathbf{5\ \dfrac{\text{miles}}{\text{hours}}}\)

1) 

The school that Carlos goes to is selling tickets to a choral performance.

On the first day of ticket sales the school sold 14 senior citizen tickets and 6 student tickets for a total of $180.

The school took in $117 on the second day by selling 12 senior citizen tickets and 1 student ticket.

Find the price of a senior citizen ticket and the price of a student ticket. 

x = senior citizen ticket price
y = student ticket price

\(\begin{array}{|lrcll|} \hline (1) & 12x+1y &=& $117 \\ & y &=& 117 - 12x \\\\ (2) & 14x+6y &=& $180 \\ & 14x+6(117 - 12x) &=& $180 \quad &| \quad :2 \\ & 7x+3(117 - 12x) &=& 90 \\ & 7x+ 351 - 36x &=& 90 \\ & 351 - 29x &=& 90 \quad &| \quad + 29x \\ & 351 &=& 90 + 29x \quad &| \quad - 90 \\ & 261 &=& 29x \quad &| \quad : 29 \\ & 9 &=& x \\ & \mathbf{x} & \mathbf{=} & \mathbf{$9} \\\\ & y &=& 117 - 12x \quad &| \quad x=9 \\ & y &=& 117 - 12(9) \\ & y &=& 117 - 108 \\ & y &=& 9 \\ & \mathbf{y} & \mathbf{=} & \mathbf{$9} \\ \hline \end{array}\)

The price of a senior citizen ticket is $9.
The price of a student ticket is $9.

Two fair 6-sided dice are rolled.

1. What is the expected value of the sum of the dice?

2. What is the expected value of the product of the dice?

1. What is the expected value of the sum of the dice?

\(\begin{array}{|r|r|r|r|r|r|r|} \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline 5 & 6 & 7 & 8 & 9 & 10& 11 \\ \hline 6 & 7 & 8 & 9 & 10& 11& 12 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{Expected value}\\ = \dfrac{1\cdot 2 + 2\cdot 3+ 3\cdot 4+ 4\cdot 5+ 5\cdot 6 + 6\cdot 7+ 5\cdot 8+ 4\cdot 9+ 3\cdot 10+ 2\cdot 11 + 1\cdot12}{36} \\ = \dfrac{252}{36} \\\\ = 7 \\ \hline \end{array}\)

2. What is the expected value of the product of the dice?

\(\begin{array}{|r|r|r|r|r|r|r|} \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 2 & 2 & 4 & 6 & 8 & 10& 12 \\ \hline 3 & 3 & 6 & 9 & 12& 15& 18 \\ \hline 4 & 4 & 8 & 12& 16& 20& 24 \\ \hline 5 & 5 & 10& 15& 20& 25& 30 \\ \hline 6 & 6 & 12& 18& 24& 30& 36 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{Expected value}\\ =\dfrac{ 6\cdot \dfrac{1+6}{2}+6\cdot \dfrac{2+12}{2} + 6\cdot \dfrac{3+18}{2}+ 6\cdot \dfrac{4+24}{2}+ 6\cdot \dfrac{5+30}{2}+ 6\cdot \dfrac{6+36}{2} } {36}\\ = \dfrac{441}{36} \\\\ = 12.25 \\ \hline \end{array}\)

Simplify

\(\large{i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.} \)

Geometric sequence:

\(\begin{array}{|rcll|} \hline s &=& i^1+&i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99} \quad & | \quad \cdot i \\ i\cdot s &=& &i^2+i^3+i^4+\cdots+ i^{98} + i^{99}+i^{100} \\\\ \hline s-i\cdot s &=& i^1 - i^{100} \\ s(1-i) &=& i^1 - i^{100} \\\\ \mathbf{s} &\mathbf{=} & \mathbf{ \dfrac{i - i^{100}}{1-i} } \quad & | \quad i^{100}= \left(i^2 \right)^{50}=(-1)^{50}=1 \\\\ s & = & \dfrac{i-1}{1-i} \\ s & = & -\left( \dfrac{1-i}{1-i} \right) \\ \mathbf{s} &\mathbf{=} & \mathbf{-1} \\ \hline \end{array}\)

\(\boxed{\large{i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99} = -1}}\)

Formel vereinfachen: 

\(\dfrac{2b}{\sqrt{a+b}-\sqrt{a-b} }\)

\(\begin{array}{|rcll|} \hline && \dfrac{2b}{\sqrt{a+b}-\sqrt{a-b} } \\\\ &=& \left(\dfrac{2b}{\sqrt{a+b}-\sqrt{a-b} } \right) \cdot \left( \dfrac{ \sqrt{a+b}+\sqrt{a-b} } { \sqrt{a+b}+\sqrt{a-b} } \right) \\\\ &=& \dfrac{2b(\sqrt{a+b}+\sqrt{a-b})}{(\sqrt{a+b}-\sqrt{a-b})(\sqrt{a+b}+\sqrt{a-b}) } \quad | \quad \text{binom: }~\boxed{(a-b)(a+b)=a^2-b^2} \\\\ &=& \dfrac{2b(\sqrt{a+b}+\sqrt{a-b})}{\left(\sqrt{a+b}\right)^2-\left(\sqrt{a-b}\right)^2 } \\\\ &=& \dfrac{2b(\sqrt{a+b}+\sqrt{a-b})}{ a+b -(a-b) } \\\\ &=& \dfrac{2b(\sqrt{a+b}+\sqrt{a-b})}{ a+b -a +b } \\\\ &=& \dfrac{2b(\sqrt{a+b}+\sqrt{a-b})}{ 2b } \\\\ &=& \sqrt{a+b}+\sqrt{a-b} \\ \hline \end{array}\)

Evaluate the sum

\(\large{\dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \cdots + \dfrac{k}{2^k} + \cdots}\)
\large{\dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \cdots + \dfrac{k}{2^k} + \cdots}

Binomial power:

\(\begin{array}{|rcll|} \hline (1-x)^{-2} &=& \dbinom{-2}{0} - \dbinom{-2}{1}x + \dbinom{-2}{2}x^2- \dbinom{-2}{3}x^3+ \dbinom{-2}{4}x^4 -\dbinom{-2}{5}x^5 \\ && +- \cdots \infty \\\\ &=& 1 -(-2)x + \dfrac{(-2)(-2-1)}{2\cdot 1}x^2 - \dfrac{(-2)(-2-1)(-2-2)}{3\cdot 2\cdot 1}x^3 \\ && +\dfrac{(-2)(-2-1)(-2-2)(-2-3)}{4\cdot 3\cdot 2\cdot 1}x^4 \\ && -\dfrac{(-2)(-2-1)(-2-2)(-2-3)(-2-4)}{5\cdot 4\cdot 3\cdot 2\cdot 1}x^5 +- \cdots \infty \\\\ &=& 1 +2x + \dfrac{(-2)(-3)}{2}x^2 - \dfrac{(-2)(-3)(-4)}{3\cdot 2}x^3 \\ && +\dfrac{(-2)(-3)(-4)(-5)}{4\cdot 3\cdot 2}x^4 -\dfrac{(-2)(-3)(-4)(-5)(-6)}{5\cdot 4\cdot 3\cdot 2}x^5 +- \cdots \infty \\\\ &=& 1 +2x + 3x^2 +4x^3 + 5x^4+6x^5 + \cdots \infty \\\\ \hline \\ x\cdot (1-x)^{-2} &=& x\cdot \left( 1 +2x + 3x^2 +4x^3 + 5x^4+6x^5 + \cdots \infty \right) \\\\ &=& x +2x^2 + 3x^3 +4x^4 + 5x^5+6x^6 + \cdots + k x^k + \cdots \infty \quad | \quad x=\dfrac12 \\\\ \dfrac12\left(1-\dfrac12 \right)^{-2} &=& \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty \\\\ \dfrac12\left(\dfrac12 \right)^{-2} &=& \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty \\\\ \dfrac12\left(\dfrac21 \right)^{2} &=& \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty \\\\ \dfrac42 &=& \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty \\\\ 2 &=& \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty \\ \hline \end{array}\)

\(\mathbf{\dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3}+ \dfrac{4}{2^4}+ \dfrac{5}{2^5}+ \dfrac{6}{2^6} + \cdots + \dfrac{k}{2^k} + \cdots \infty=2 } \\\)

If a + b = 9 and x + y = 1,

what is -3b - 3y - 3a - 3x?

\(\begin{array}{|lrcll|} \hline & a+b &=& 9 \\ & x+y &=& 1 \\ \text{sum} & a+b+x+y &=& 9+1 \\ & a+b+x+y &=& 10 \\\\ & && -3b - 3y - 3a - 3x \\ & &=& -3\cdot (b+y+a+x) \\ & &=& -3\cdot (a+b+x+y) \quad & | \quad a+b+x+y = 10 \\ & &=& -3\cdot (10) \\ & &=& -30 \\ \hline \end{array}\)

Please help!

Solve the system of equations by elimination:

\(\begin{array}{|lrcll|} \hline (1) & 4x + 8y &=& -16 \quad & | \quad -8y \\ & 4x &=& -16 -8y \\\\ (2) & -8x + 10y &=& -20 \\ & -2\cdot (4x) + 10y &=& -20 \quad & | \quad 4x = -16 -8y \\ & -2\cdot ( -16 -8y) + 10y &=& -20 \\ & 32 + 16y + 10y &=& -20 \\ & 32 + 26y &=& -20 \quad & | \quad -32 \\ & 26y &=& -20-32 \\ & 26y &=& -52 \quad & | \quad : 26 \\ & y &=& -\frac{52}{26} \\ & \mathbf{y} & \mathbf{=} & \mathbf{-2} \\\\ & 4x &=& -16 -8y \quad & | \quad :4 \\ & x &=& -4 -2y \quad & | \quad \mathbf{y = -2} \\ & x &=& -4 -2\cdot (-2) \\ & x &=& -4 +4 \\ & \mathbf{x} & \mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

When the positive integers are arranged in order,
filling in the successive diagonals of an infinite grid from top to bottom, as shown,
the integer 41 is in the (5,5) spot.
What integer would we see in the (10,10) spot if the rest of the grid were visible?

\(\text{The red entries in the main diagonal (n,n) form} \\ \text{an arithmetic series of the second order: {$1,5,13,25,41,\ldots$}}\)

\(\begin{array}{|r|r|r|r|} \hline n & (n,n ) & & \text{First difference} & \text{Second difference} \\ \hline 1 & (1,1) & 1 & \\ & & & 4 \\ 2 & (2,2) & 5 & & 4 \\ & & & 8 \\ 3 & (3,3) & 13 & & 4 \\ & & & 12 \\ 4 & (4,4) & 25 & & 4 \\ & & & 16 \\ 5 & (5,5) & 41 & & 4 \\ & & & 20 \\ 6 & (6,6) & 61 & & 4 \\ & & & 24 \\ 7 & (7,7) & 85 & & 4 \\ & & & 28 \\ 8 & (8,8) & 113 & & 4 \\ & & & 32 \\ 9 & (8,9) & 145 & & 4 \\ & & & 36 \\ 10 & (10,10) & \mathbf{181} & & 4 \\ \hline \end{array}\)

Compute the sum of

\(\mathbf{\dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \cdots}\)

\mathbf{\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots}