heureka

Let us have four distinct collinear points A, B, C and D on the Cartesian plane.

The point C is such that AB/CB =1/2 and

the point D is such that DA/BA = 3 and

DB/BA = 2.

If C = (0, 4), D = (4, 0) and A = (x,y) what is the value of 2x+y?

\(\text{Let $AB=BA$} \\ \text{Let $DB=BD$} \\ \text{Let $\vec{C}=\dbinom{0}{4}$ } \\ \text{Let $\vec{D}=\dbinom{4}{0}$ }\)

\(\begin{array}{|rcll|} \hline \dfrac{AB}{CB} &=& \dfrac{1}{2} \\ \mathbf{ CB } &=& \mathbf{2AB} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{DB}{BA} &=& 2\\ \mathbf{ DB } &=& \mathbf{2AB} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline CD &=& CB + BD \\ CD &=& 2AB+2AB \\ \mathbf{ CD } &=& \mathbf{4AB} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{DA}{BA} &=& 3 \\ \mathbf{ DA } &=& \mathbf{3AB} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{DA}{CD} &=& \dfrac{3AB}{4AB} \\ \mathbf{\lambda_A=\dfrac{DA}{CD} }&=& \mathbf{\dfrac{3}{4}} \\ \hline \end{array}\)

Line:

\(\begin{array}{|rcll|} \hline \vec{x} &=& \vec{D}+\lambda(\vec{C}-\vec{D}) \\ \vec{A} &=& \vec{D}+\lambda_A(\vec{C}-\vec{D}) \quad | \quad \lambda_A = \dfrac{3}{4} \\ \vec{A} &=& \vec{D}+\dfrac{3}{4}(\vec{C}-\vec{D}) \\ &=& \dbinom{4}{0}+\dfrac{3}{4}\left( \dbinom{0}{4}-\dbinom{4}{0} \right) \\\\ &=& \dbinom{4}{0}+\dfrac{3}{4} \dbinom{-4}{4} \\\\ &=& \dbinom{4}{0}+ \dbinom{\dfrac{3}{4}\cdot(-4)}{\dfrac{3}{4}\cdot 4} \\\\ &=& \dbinom{4}{0}+ \dbinom{-3} {3} \\\\ &=& \dbinom{4-3}{3} \\\\ \mathbf{\vec{A} } &=& \mathbf{ \dbinom{1}{3} } \\\\ \vec{A} &=& \dbinom{x}{y} \qquad x = 1,\ y=3 \\ 2x+y &=& 2\cdot 1 + 3 \\ \mathbf{2x+y } &=& \mathbf{ 5 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{DB}{CD} &=& \dfrac{2AB}{4AB} \\ \mathbf{\lambda_B=\dfrac{DB}{CD} }&=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)

Line:

\(\begin{array}{|rcll|} \hline \vec{x} &=& \vec{D}+\lambda(\vec{C}-\vec{D}) \\ \vec{B} &=& \vec{D}+\lambda_B(\vec{C}-\vec{D}) \quad | \quad \lambda_B = \dfrac{1}{2} \\ \vec{B} &=& \vec{D}+\dfrac{1}{2}(\vec{C}-\vec{D}) \\ &=& \dbinom{4}{0}+\dfrac{1}{2}\left( \dbinom{0}{4}-\dbinom{4}{0} \right) \\\\ &=& \dbinom{4}{0}+\dfrac{1}{2} \dbinom{-4}{4} \\\\ &=& \dbinom{4}{0}+ \dbinom{\dfrac{1}{2}\cdot(-4)}{\dfrac{1}{2}\cdot 4} \\\\ &=& \dbinom{4}{0}+ \dbinom{-2} {2} \\\\ &=& \dbinom{4-2}{2} \\\\ \mathbf{\vec{B} } &=& \mathbf{ \dbinom{2}{2} } \\ \hline \end{array}\)

Evaluate the sum

\(\large{s= 1 + \dfrac{3}{3} + \dfrac{5}{9} + \dfrac{7}{27} + \dfrac{9}{81} + \dotsb }\)

\(\begin{array}{|rcll|} \hline s &=& 1 + \dfrac{3}{3} + \dfrac{5}{9} + \dfrac{7}{27} + \dfrac{9}{81} + \dotsb \\\\ &=& \dfrac{1}{3^0} + \dfrac{3}{3^1} + \dfrac{5}{3^2} + \dfrac{7}{3^3} + \dfrac{9}{3^4} + \dotsb + \dfrac{2n-1}{3^{n-1}} + \dotsb \\\\ \hline s_n &=& \dfrac{1}{3^0} + \dfrac{3}{3^1} + \dfrac{5}{3^2} + \dfrac{7}{3^3} + \dfrac{9}{3^4} + \dotsb + \dfrac{2n-1}{3^{n-1}} \\\\ &=& 1\cdot 3^0 + 3\cdot 3^{-1} + 5\cdot 3^{-2} + 7\cdot 3^{-3} + 9\cdot 3^{-4} + \dotsb + (2n-1)\cdot 3^{1-n} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_n &=& 1\cdot 3^0 + & 3\cdot 3^{-1} + 5\cdot 3^{-2} + 7\cdot 3^{-3} + 9\cdot 3^{-4} + \dotsb + (2n-1)\cdot 3^{-n+1} \\ 3^{-1} s_n &=& & 1\cdot 3^{-1} + 3\cdot 3^{-2} + 5\cdot 3^{-3} + 7\cdot 3^{-4} + \dotsb + (2n-3)\cdot 3^{-n+1}+ (2n-1)\cdot 3^{-n} \\ \hline \end{array} \\ \begin{array}{rcll} s_n-3^{-1} s_n &=& 1\cdot 3^0 +2\cdot ( 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1}) - (2n-1)\cdot 3^{-n} \\ s_n(1-3^{-1}) &=& 1 +2\cdot ( \underbrace{3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1}}_{\text{geometric progression}} ) - (2n-1)\cdot 3^{-n} \\ \dfrac{2}{3} s_n &=& 1 +2\cdot \mathbf{ S_n } - (2n-1)\cdot 3^{-n} \\ \end{array}\\ \begin{array}{|rcll|} \hline S_n &=& 3^{-1} + & 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1} \\ 3^{-1} S_n &=& & 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1} + 3^{-n} \\ \hline \end{array} \\ \begin{array}{rcll} S_n-3^{-1} S_n &=& 3^{-1} - 3^{-n} \\ S_n(1-3^{-1} ) &=& 3^{-1} - 3^{-n} \\ \dfrac{2}{3} S_n &=& 3^{-1} - 3^{-n} \\ \mathbf{ S_n } &=& \mathbf{\dfrac{3}{2} ( 3^{-1} - 3^{-n} )} \\ \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{2}{3} s_n &=& 1 +2\cdot \mathbf{ S_n } - (2n-1)\cdot 3^{-n} \quad | \quad \mathbf{ S_n = \dfrac{3}{2} ( 3^{-1} - 3^{-n} )} \\\\ \dfrac{2}{3} s_n &=& 1 +3\cdot ( 3^{-1} - 3^{-n} ) - (2n-1)\cdot 3^{-n} \\\\ \dfrac{2}{3} s_n &=& 1 +3^1 3^{-1} - 3^13^{-n} - (2n-1)\cdot 3^{-n} \\\\ \dfrac{2}{3} s_n &=& 1 + 1 - 3^{-n}(3 + 2n-1 ) \\\\ \dfrac{2}{3} s_n &=& 2- 3^{-n}(2n+2) \\\\ \dfrac{2}{3} s_n &=& 2- 2\cdot 3^{-n}(n+1) \\\\ \dfrac{2}{3} s_n &=& 2\Big(1- 3^{-n}(n+1)\Big) \quad | \quad : 2 \\\\ \dfrac{1}{3} s_n &=& 1- 3^{-n}(n+1) \\ s_n &=& 3\Big(1- 3^{-n}(n+1)\Big) \\ &=& 3\Big(1- \dfrac{n+1} {3^{n} }\Big) \\ &=& 3\Big(\dfrac{3^{n}-n-1} {3^{n} }\Big) \\ &=& \dfrac{3^{n}-n-1} {3^{-1}\cdot 3^{n} } \\ \mathbf{ s_n } &=& \mathbf{\dfrac{3^{n}-n-1} {3^{n-1} } } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s = \sum \limits_{n=1}^{\infty}\dfrac{2n-1}{3^{n-1}} = \lim \limits_{n\to\infty} s_n &=& \lim \limits_{n\to\infty}\dfrac{3^{n}-n-1} {3^{n-1} } \quad | \quad \text{l'Hospital's rule} \\ &=& \lim \limits_{n\to\infty}\dfrac{n3^{n-1}-1} {(n-1)3^{n-2} } \quad | \quad \text{l'Hospital's rule} \\ &=& \lim \limits_{n\to\infty}\dfrac{n(n-1)3^{n-2}} {(n-1)(n-2)3^{n-3} } \\\\ &=& \lim \limits_{n\to\infty}\dfrac{n(n-1)3^{n-2}} {(n-1)(n-2)3^{n-2}3^{-1} } \\\\ &=& \lim \limits_{n\to\infty}\dfrac{n} { (n-2)3^{-1} } \\\\ &=& 3\cdot \lim \limits_{n\to\infty}\dfrac{n} {n-2} \\\\ &=& 3\cdot \lim \limits_{n\to\infty}\dfrac{1} {\dfrac{n-2}{n} } \\\\ &=& 3\cdot \lim \limits_{n\to\infty}\dfrac{1} {1-\dfrac{2}{n} } \quad | \quad \lim \limits_{n\to\infty} \dfrac{2}{n} = 0 \\\\ &=& 3 \cdot \dfrac{1} {1-0 } \\\\ &=& 3 \cdot \dfrac{1} {1 } \\\\ \mathbf{ s } &=& \mathbf{3} \\ \hline \end{array}\)

I am supposed to find all of the solutions to the equation below in the interval [0, 360).

Round approximate answers to the nearest tenth of a degree. 

The equation is: sin(x) - cos(x) = 0.8

What's the most efficient way to solve this? 

I.

Formula:

\(\begin{array}{|rcll|} \hline \sin(x-45^\circ) &=& \sin(x)\cos(45^\circ) - \cos(x)\sin(45^\circ) \\\\ && \qquad \boxed{\sin(45^\circ)=\cos(45^\circ)=\dfrac{\sqrt{2}} {2}} \\\\ \sin(x-45^\circ) &=& \sin(x)\dfrac{\sqrt{2}} {2} - \cos(x)\dfrac{\sqrt{2}} {2} \\\\ \sin(x-45^\circ) &=& \dfrac{\sqrt{2}} {2} \Big(\sin(x) - \cos(x) \Big) \\\\ \mathbf{\sin(x) - \cos(x)} &\mathbf{=}& \mathbf{\dfrac{2} {\sqrt{2}}\sin(x-45^\circ)} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sin(x) - \cos(x) &=& 0.8 \quad | \quad \mathbf{\sin(x) - \cos(x) = \dfrac{2} {\sqrt{2}}\sin(x-45^\circ)} \\ \dfrac{2} {\sqrt{2}}\sin(x-45^\circ) &=& 0.8 \\ \sin(x-45^\circ) &=& 0.8 \dfrac{\sqrt{2}}{2} \\ \mathbf{\sin(x-45^\circ)} &\mathbf{=}& \mathbf{0.4\sqrt{2}} \\ \hline \end{array} \)

Solution 1:

\(\begin{array}{|rcll|} \hline \mathbf{\sin(x-45^\circ)} &\mathbf{=}& \mathbf{0.4\sqrt{2}} \quad | \quad \arcsin() \\ x-45^\circ & = & \arcsin(0.4\sqrt{2})+ 360z,\ z \in \mathbb{Z} \\ x & = & 45^\circ + \arcsin(0.4\sqrt{2})+ 360z \\ & = & 45^\circ + \arcsin(0.56568542495)+ 360z \\ & = & 45^\circ + 34.4499019880^\circ + 360z \\ & = & 79.4499019880^\circ + 360z \\ & = & 79.4499019880^\circ \quad | \quad \text{interval [0, 360)} \\ \mathbf{ x } & \mathbf{=} & \mathbf{79.4^\circ} \\ \hline \end{array} \)

Solution 2:

\(\begin{array}{|rcll|} \hline \mathbf{\sin(x-45^\circ)} &\mathbf{=}& \mathbf{0.4\sqrt{2}} \\\\ && \boxed{\sin(x-45^\circ) = \sin\Big(180^\circ-(x-45^\circ)\Big) \\ \sin(x-45^\circ) = \sin(180^\circ-x+45^\circ) \\ \mathbf{\sin(x-45^\circ) = \sin(225^\circ-x)} } \\\\ \mathbf{\sin(225^\circ-x)} &\mathbf{=}& \mathbf{0.4\sqrt{2}} \quad | \quad \arcsin() \\ 225^\circ-x & = & \arcsin(0.4\sqrt{2})+ 360z,\ z \in \mathbb{Z} \\ x &=& 225^\circ - \arcsin(0.4\sqrt{2})+ 360z \\ &=& 225^\circ - \arcsin(0.56568542495)+ 360z \\ &=& 225^\circ - 34.4499019880^\circ + 360z \\ &=& 190.550098012^\circ + 360z \quad | \quad \text{interval [0, 360)} \\ &=& 190.550098012^\circ \\ \mathbf{ x } & \mathbf{=} & \mathbf{190.6^\circ} \\ \hline \end{array}\)

Let \(x,y,z\) be real numbers such that

\(\begin{align*} x + y + z &= 4, \\ x^2 + y^2 + z^2 &= 6. \end{align*}\)
Let \(m\) and \(M\) be the smallest and largest possible values of \(x\) respectively.

Find \(m+M\)

Easy Golden Ratio Quadratic

\(\dfrac{l+w}{l}=\dfrac{l}{w}\)

I assume:

\(\text{Let }\textbf{Golden Ratio} =\varphi\)

\(\begin{array}{|rcll|} \hline \dfrac{l+w}{l} &=& \dfrac{l}{w} \\\\ \dfrac{l}{l}+ \dfrac{w}{l} &=& \dfrac{l}{w} \\\\ 1 + \dfrac{w}{l} &=& \dfrac{l}{w} \quad | \quad \dfrac{l}{w} = \varphi,\ \dfrac{w}{l} = \dfrac{1}{\varphi} \\\\ 1 + \dfrac{1}{\varphi} &=& \varphi \quad |\quad \cdot \varphi \\\\ \varphi + 1 &=& \varphi^2 \\ \varphi^2 -\varphi - 1 &=& 0 \\\\ \varphi &=& \dfrac{1\pm \sqrt{1-4\cdot (-1) }}{2} \\ \varphi &=& \dfrac{1\pm \sqrt{5}}{2} \\\\ \mathbf{\varphi} &\mathbf{=}& \mathbf{\dfrac{1+ \sqrt{5}}{2}} \\ \hline \end{array}\)

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

 \(\begin{array}{|rcll|} \hline && \displaystyle i^1+i^2+i^3+i^4+i^5+\cdots+ i^{97} + i^{98}+i^{99} \\\\ &=& (i^1+i^2+i^3+i^4)(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \\\\ &=& (i^1+i^2)(1+i^2)(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \quad | \quad i^2 = -1 \\\\ &=& (i^1+i^2)(1-1)(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \\\\ &=& (i^1+i^2)(\underbrace{1-1}_{=0})(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \\\\ &=& 0-i^{100} \\\\ &=& -(i^2)^{50} \quad | \quad i^2 = -1 \\\\ &=& -(-1)^{50} \\\\ &=& -1 \\ \hline \end{array}\)

 Trigonometric Identities:

\(\large\dfrac{2\tan(x)}{1-\tan^2(x)} = 1\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{2\tan(x)}{1-\tan^2(x)} } \\\\ &=& \dfrac{ \tan(x)+\tan(x)}{1-\tan(x)\tan(x)} \quad | \quad \boxed{ \text{Formula: } \tan(x+y) = \dfrac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} } \\\\ &\mathbf{=}& \mathbf{\tan(2x)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \tan(2x) &=& 1 \\ 2x &=& \arctan(1) + n\pi,\ n \in \mathbb{Z} \quad | \quad \arctan(1) = \dfrac{\pi}{4} \\ 2x &=& \dfrac{\pi}{4} + n\pi \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{\pi}{8} + n\dfrac{\pi}{2}} \\ \hline \end{array}\)

\(\text{Split } \mathbf{\dfrac{\pi}{8} + n\dfrac{\pi}{2}}\text{ in two solutions:}\)

\(\begin{array}{|rcll|} \hline x & = & \dfrac{\pi}{8} + n\dfrac{\pi}{2} \\ x_1 &=& \dfrac{\pi}{8} + n\left(\dfrac{\pi}{2}+\dfrac{\pi}{2} \right) \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\dfrac{\pi}{8} + n\pi} \\\\ x_2 &=& \dfrac{\pi}{8}+\dfrac{\pi}{2} + n\left(\dfrac{\pi}{2}+\dfrac{\pi}{2} \right) \\ \mathbf{x_2} &\mathbf{=}& \mathbf{\dfrac{5\pi}{8} + n\pi} \\ \hline \end{array}\)

 Please solve the whole problem fully

For each sequence, decide whether it could be arithmetic, geometric, or neither.

\(\begin{array}{|l|l|c|} \hline 1. & 25,5,1,\ldots & \text{geometric} \quad r=\dfrac{1}{5} \\ \hline 2. & 25,19,13,\ldots & \text{arithmetic} \quad d=-6 \\ \hline 3. & 4,9,16,\ldots \\ & 4,10,16,\ldots & \text{arithmetic} \quad d=6 \\ & 4,8,16,\ldots & \text{geometric} \quad r=2 \\ \hline 4. & 50,60,70,\ldots & \text{arithmetic} \quad d=10 \\ \hline 5. & \dfrac{1}{2},3,18,\ldots & \text{geometric} \quad r = 6 \\ \hline \end{array}\)

Express the following in simplest form : \((-2)^{4x+2}(2)^{6x-5}(8)^x\)

\(\begin{array}{|rcll|} \hline && \mathbf{(-2)^{4x+2}(2)^{6x-5}(8)^x} \\ &=& \Big((-1)(2) \Big)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad \boxed{(ab)^c = a^cb^c} \\ &=& \left(-1 \right)^{4x+2}\left(2 \right)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad \boxed{a^{b+c} = a^ba^c} \\ &=& \left(-1 \right)^{4x}\left(-1 \right)^{2}\left(2 \right)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad \boxed{ (-1)^2 = (-1)(-1)=1 } \\ &=& \left(-1 \right)^{4x}\left(2 \right)^{4x+2}(2)^{6x-5}(8)^x \quad & | \quad 8 = 2^3 \\ &=& \left(-1 \right)^{4x}\left( 2 \right)^{4x+2}(2)^{6x-5} \left(2^3 \right)^x \quad & | \quad \boxed{ \left(a^b \right)^c = a^{bc} \\ \left(2^3 \right)^x = 2^{3x} } \\ &=& \left(-1 \right)^{4x}\left( 2 \right)^{4x+2}(2)^{6x-5} (2^{3x}) \quad & | \quad \boxed{ a^ba^ca^d=a^{b+c+d} } \\ &=& \left(-1 \right)^{4x}2^{4x+2+6x-5+3x} \\ &=& \left(-1 \right)^{4x}2^{4x+2+6x-5+3x} \\ &\mathbf{=}& \mathbf{\left(-1 \right)^{4x}2^{13x-3}} \\ \hline \end{array} \)