heureka

how to work out the size of an exterior angle of a regular pentagon

The second one for today is also about solving

\(\mathbf{7^{x^2-9}=1}\)

\(\begin{array}{|rcll|} \hline 7^{x^2-9} &=& 1 \quad & | \quad \cdot 7 \\ 7^{x^2-9}\cdot 7 &=& 7 \\ 7^{x^2-9}\cdot 7^1 &=& 7 \quad & | \quad a^b * a^c = a^{b+c} \\ 7^{x^2-9+1} &=& 7 \\ 7^{x^2-8} &=& 7^1 \\ x^2-8 &=& 1 \\ x^2 &=& 9 \quad & | \quad \text{square root both sides} \\ x &=& \pm 3 \\ \hline \end{array}\)

determining the sum of this arithmetic series?

5 + 8 + 11 + ... + 53

arithmetic series:

\(\begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 5} && 8 && 11 && 14 && 17 && \cdots && 53 \\ \text{1. Difference } && {\color{red}d_1 = 3} && 3 && 3 && 3 && \cdots && 3 \\ \end{array}\)

Formula:

\(\begin{array}{|rcl|} \hline a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } \\ s &=& \binom{n}{1}\cdot {\color{red}d_0 } + \binom{n}{2}\cdot {\color{red}d_1 } \\ \hline \end{array}\)

n = ?

\(\begin{array}{|rcl|} \hline a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } \quad & | \quad a_n = 53 \quad {\color{red}d_0 } = 5 \quad {\color{red}d_1 } = 3 \\\\ 53 &=& \binom{n-1}{0}\cdot {\color{red}5 } + \binom{n-1}{1}\cdot {\color{red}3 } \\ 53 &=& 5+(n-1)\cdot 3 \\ 53 &=& 2 +3n \quad & | \quad -2\\ 51 &=& 3n \quad & | \quad :3 \\ 17 &=& n \\ \mathbf{n} &\mathbf{=}& \mathbf{17} \\ \hline \end{array} \)

sum (s) = ?

\(\begin{array}{|rcl|} \hline s &=& \binom{n}{1}\cdot {\color{red}d_0 } + \binom{n}{2}\cdot {\color{red}d_1 } \quad & | \quad n = 17 \quad {\color{red}d_0 } = 5 \quad {\color{red}d_1 } = 3 \\\\ s &=& \binom{17}{1}\cdot {\color{red}5 } + \binom{17}{2}\cdot {\color{red}3 } \\ s &=& 17\cdot 5 + \frac{17}{2} \cdot \frac{16}{1}\cdot 3 \\ s &=& 17\cdot \left(5+\frac{16}{2}\cdot 3 \right) \\ s &=& 17\cdot(5+ 24) \\ s &=& 17\cdot 29 \\ \mathbf{s} &\mathbf{=}& \mathbf{493} \\ \hline \end{array}\)

The sum is 493

Find a complex number z such that the real part and imaginary part of z are both integers,

and such that zz = 89

I assume, \( z\bar{z} = 89\)

89 has one solution, because 89 is a prime number and \(89 = 1 \pmod{4}\)

\(89 = 5^2 + 8^2\)

so

\(z = 5 + 8i \text{ and } \bar{z} = 5-8i \text{ so } z\bar{z} = 5^2+8^2 = 89\\ \text{or} \\ z= 8 + 5i \text{ and } \bar{z} = 8-5i \text{ so } z\bar{z} = 8^2+5^2 = 89\)

Express 1/(1+1/(1-1/(1-i))) in the form a+bi

where a and b are real numbers.

\(\begin{array}{|rcll|} \hline && \mathbf{\cfrac{1}{1+\cfrac{1}{1-\cfrac{1}{ 1-i }}}} \\\\ &=& \cfrac{1}{1+\cfrac{1}{\cfrac{1-i}{ 1-i }-\cfrac{1}{ 1-i }}} \\\\ &=& \cfrac{1}{1+\cfrac{1}{\cfrac{1-i-1}{ 1-i }}} \\\\ &=& \cfrac{1}{1+\cfrac{1-i}{ \not{1}-i-\not{1} }} \\\\ &=& \cfrac{1}{1+\cfrac{1-i}{ -i} } \\\\ &=& \cfrac{1}{\cfrac{-i}{-i}+\cfrac{1-i}{-i} } \\\\ &=& \cfrac{1}{\cfrac{-i+1-i}{-i} } \\\\ &=& \dfrac{-i}{-i+1-i} \\\\ &=& \dfrac{-i}{1-2i} \\\\ &=& \left(\dfrac{-i}{1-2i} \right) \cdot \left( \dfrac{1+2i}{1+2i} \right) \\\\ &=& \dfrac{-i(1+2i)}{(1-2i)(1+2i)} \\\\ &=& \dfrac{-i-2i^2}{1-4i^2} \quad & | \quad i^2 = -1 \\\\ &=& \dfrac{-i-2(-1)}{1-4(-1)} \\\\ &=& \dfrac{-i+2}{1+4} \\\\ &=& \dfrac{-i+2}{5} \\\\ & \mathbf{=}& \mathbf{ \dfrac{2}{5}-\dfrac{1}{5}i } \\ \hline \end{array}\)

The system of equations \[\frac{3xy}{x + y} = 5, \quad \frac{2xz}{x + z} = 3, \quad \frac{yz}{y + z} = 4\]

has one ordered triple solution $(x,y,z)$.

What is the value of $z$ in this solution? 

\(\begin{array}{|rcll|} \hline \dfrac{3xy}{x + y} = 5, \quad \dfrac{2xz}{x + z} = 3, \quad \dfrac{yz}{y + z} = 4 \\ z=\ ? \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \dfrac{3xy}{x + y} &=& 5 \\\\ & \dfrac{3}{5} &=& \dfrac{x+y}{xy} \\\\ \mathbf{(1)} & \mathbf{\dfrac{3}{5}} &\mathbf{=}& \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \\\\ \hline & \dfrac{2xz}{x + z} &=& 3 \\\\ & \dfrac{2}{3} &=& \dfrac{x+z}{xz} \\\\ \mathbf{(2)} & \mathbf{ \dfrac{2}{3}} &\mathbf{=}& \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \\\\ \hline & \dfrac{yz}{y + z} &=& 4 \\ & \dfrac{1}{4} &=& \dfrac{y+z}{yz} \\\\ \mathbf{(3)} & \mathbf{\dfrac{1}{4}} &\mathbf{=}& \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} \\ \hline \end{array}\)

z = ?

\(\begin{array}{|lrcll|} \hline (2)+(3)-(1): & \mathbf{ \dfrac{2}{3}}+\mathbf{\dfrac{1}{4}} - \mathbf{\dfrac{3}{5}} &=& \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \right) + \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} \right) - \left( \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \right) \\\\ & \dfrac{2}{3} +\dfrac{1}{4} - \dfrac{3}{5} &=& \dfrac{2}{z} \\\\ & \dfrac{2}{z} &=& \dfrac{2}{3} +\dfrac{1}{4} - \dfrac{3}{5} \\\\ & \dfrac{2}{z} &=& \dfrac{2\cdot 20+15-3\cdot 12}{60} \\\\ & \dfrac{1}{z} &=& \dfrac{40+15-36}{120} \\\\ & \dfrac{1}{z} &=& \dfrac{19}{120} \\\\ & \mathbf{z} &\mathbf{=}& \mathbf{\dfrac{120}{19}} \\ \hline \end{array}\)

x = ?

\(\begin{array}{|lrcll|} \hline (1)-(3)+(2): & \mathbf{\dfrac{3}{5}}-\mathbf{\dfrac{1}{4}}+\mathbf{ \dfrac{2}{3}} &=& \left( \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \right) - \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} \right) + \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \right) \\\\ & \dfrac{3}{5} -\dfrac{1}{4} + \dfrac{2}{3} &=& \dfrac{2}{x} \\\\ & \dfrac{2}{x} &=& \dfrac{3}{5} -\dfrac{1}{4} + \dfrac{2}{3} \\\\ & \dfrac{2}{x} &=& \dfrac{ 3\cdot 12-15+2\cdot 20}{60} \\\\ & \dfrac{1}{x} &=& \dfrac{36-15+40}{120} \\\\ & \dfrac{1}{x} &=& \dfrac{61}{120} \\\\ & \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{120}{61}} \\ \hline \end{array}\)

y = ?

\(\begin{array}{|lrcll|} \hline (1)+(3)-(2): & \mathbf{\dfrac{3}{5}}+\mathbf{\dfrac{1}{4}}-\mathbf{ \dfrac{2}{3}} &=& \left( \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} \right) + \left( \mathbf{\dfrac{1}{z} + \dfrac{1}{y}} \right) - \left( \mathbf{\dfrac{1}{z}+\dfrac{1}{x}} \right) \\\\ & \dfrac{3}{5} +\dfrac{1}{4} - \dfrac{2}{3} &=& \dfrac{2}{y} \\\\ & \dfrac{2}{y} &=& \dfrac{3}{5} +\dfrac{1}{4} - \dfrac{2}{3} \\\\ & \dfrac{2}{y} &=& \dfrac{ 3\cdot 12+15-2\cdot 20}{60} \\\\ & \dfrac{1}{y} &=& \dfrac{36+15-40}{120} \\\\ & \dfrac{1}{y} &=& \dfrac{11}{120} \\\\ & \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{120}{11}} \\ \hline \end{array}\)

Find all complex numbers z satisfying the equation (z+1)/(z-1) = i

\(\begin{array}{|lrcll|} \hline & \dfrac{z+1}{z-1} &=& i \quad & | \quad z = a+bi \\\\ & \dfrac{a+bi+1}{a+bi-1} &=& i \\\\ & a+bi+1 &=& i(a+bi-1) \\\\ & a+bi+1 - i(a+bi-1)&=& 0 \\\\ & a+bi+1 - ia-bi^2+i&=& 0 \quad & | \quad i^2 = -1\\ \\ & a+bi+1 - ia+b+i&=& 0 \\ \\ & a+bi+1 - ia+b+i&=& 0 \\ \\ & ( a+b+1) + (b-a+1) i &=& 0 \\ \\ \hline (1) & (a+b+1) &=& 0 \\ (2) & (b-a+1) &=& 0 \\ \hline (1) + (2) & 2b+2 &=& 0 \quad & | \quad : 2 \\ & b+1 &=& 0 \\ & \mathbf{b} &\mathbf{=}& \mathbf{-1} \\ \hline (1) - (2) & 2a &=& 0 \quad & | \quad : 2 \\ & \mathbf{a} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)

\(z = a+bi \\ z = 0 -i \\ \mathbf{z = -i}\)

Simplify (i+1)^{3200}-(i-1)^{3200}

\(\begin{array}{|rclrcl|} \hline && \mathbf{(i+1)^{3200}-(i-1)^{3200}} \\ &=& (i+1)^{2\cdot 1600}-(i-1)^{2\cdot 1600} \\ &=& \left( (i+1)^{2} \right)^{1600}- \left( (i-1)^{2} \right)^{1600} \quad & |\quad (i+1)^{2} &=& i^2+2i+1 \qquad i^2 = -1\\ && \quad & \quad &=& -1+2i+1 \\ && \quad & \quad &=& 2i \\ &=& (2i)^{1600} - \left( (i-1)^{2} \right)^{1600} \quad & |\quad (i-1)^{2} &=& i^2-2i+1 \qquad i^2 = -1\\ && \quad & \quad &=& -1-2i+1 \\ && \quad & \quad &=& -2i \\ &=& (2i)^{1600} - (-2i)^{1600} \\ &=& (2i)^{1600} - (-1)^{1600}(2i)^{1600} \quad & \quad (-1)^{1600} = 1 \\ &=& (2i)^{1600} - (2i)^{1600} \\ &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)

I know how to factor a trinomial into a product of two binomials.

I know how to do the foil method.

But how do I factor an equation like this!?

(m-5)(m+4)+8

\(\begin{array}{|rcll|} \hline && \mathbf{(m-5)(m+4)+8} \\ &=& (m-5)(m+4)+4+4 \\ &=& (m-5)(m+4)+4+4 +m-m\\ &=& (m-5)(m+4)+(m+4)-(m-4) \\ &=& (m+4)[(m-5)+1]-(m-4) \\ &=& (m+4)(m-4)-(m-4) \\ &=& (m-4)[(m+4)-1] \\ &\mathbf{=}& \mathbf{(m-4)(m+3)} \\ \hline \end{array}\)

Consider the function f(x)=x³+2x²-3

A. Graph the function

B. What are the x- and y-intercepts of the graph?

x_intercept = 1

y_intercept = -3