heureka

What is the GCF and LCM of 34 and 784? 

\(\begin{array}{|rcll|} \hline 34 &=& 2^1\cdot 17^1 \cdot (7^0) \\ 784 &=& 2^4\cdot 7^2 \cdot (17^0) \\\\ && \begin{array}{|rcll|} \hline \text{gcf}(34,784) &=& 2^1\cdot 17^0 \cdot 7^0 \quad & | \quad \text{min exponent} \\ &=& 2\cdot 1 \cdot 1 \\ &=& \mathbf{2} \\\\ \text{lcm}(34,784) &=& 2^4\cdot 17^1 \cdot 7^2 \quad & | \quad \text{max exponent} \\ &=& 16\cdot 17 \cdot 49 \\ &=& \mathbf{13328} \\ \hline \end{array}\\ \hline \end{array}\)

Simplify the following expression.
Write the answer without negative exponents and use only fractions if necessary (no decimals).

\(\begin{array}{|rcll|} \hline && \dfrac{12^{a-1}b^{10}}{-3a^{-9}b^{-4}} \\\\ &=& - \dfrac{12^{a-1}b^{10}}{3a^{-9}b^{-4}} \\\\ &=& - \dfrac{12^{a}12^{-1}b^{10}}{3a^{-9}b^{-4}} \\\\ &=& - \dfrac{12^{a}b^{10}}{3\cdot 12^1a^{-9}b^{-4}} \\\\ &=& - \dfrac{12^{a}b^{10}}{36a^{-9}b^{-4}} \\\\ &=& - \dfrac{12^{a}b^{10}a^{9}b^{4}}{36} \\\\ &=& - \dfrac{12^{a}a^{9}b^{10}b^{4}}{36} \\\\ &=& - \dfrac{12^{a}a^{9}b^{10+4}}{36} \\\\ &=& - \dfrac{12^{a}a^{9}b^{14}}{36} \\ \hline \end{array} \)

Solve the following equation for x on the interval [0, 2pi]

2sin^3(4x)-sin(4x)=2cos^2(4x)-1

\(\begin{array}{|rcll|} \hline \mathbf{2\sin^3(4x)-\sin(4x)} &\mathbf{=}& \mathbf{2\cos^2(4x)-1} \\\\ \sin(4x)\Big(2\sin^2(4x)-1 \Big) &=& 2\cos^2(4x)-1 \\ -\sin(4x)\Big(1-2\sin^2(4x)\Big) &=& 2\cos^2(4x)-1 \\\\ && \boxed{\cos(8x)= 2\cos^2(4x)-1 } \\ && \boxed{\cos(8x)= 1-2\sin^2(4x) } \\\\ \mathbf{ -\sin(4x)\cos(8x) } & \mathbf{=} & \mathbf{\cos(8x)} \\ \hline \end{array}\)

Solution: \(\mathbf{ -\sin(4x)\cdot 0 = 0}\)

\(\begin{array}{|rcll|} \hline \mathbf{\cos(8x)} &\mathbf{=}& \mathbf{0} \\\\ 8x &=& \arccos(0)+\pi n,\qquad \mathbf{n \in Z} \\\\ 8x &=& -\dfrac{\pi}{2}+\pi n \quad & | \quad : 8 \\\\ x & = & \dfrac{\pi n}{8}-\dfrac{\pi}{16} \\\\ x & = & \pi\left( \dfrac{ n}{8}-\dfrac{1}{16} \right) \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\pi\left( \dfrac{ 2n-1}{16}\right)} \\ \hline \end{array} \)

Solution: \(\mathbf{ -(-1)\cos(8x) = \cos(8x) }\)

\(\begin{array}{|rcll|} \hline \mathbf{\sin(4x)} &\mathbf{=}& \mathbf{-1} \\\\ 4x &=& \arcsin(-1)+2\pi n,\qquad \mathbf{n \in Z} \\\\ 4x &=& \dfrac{3\pi}{2} + 2\pi n \quad & | \quad : 4 \\\\ x &=& \dfrac{3\pi}{8} + \dfrac{\pi n}{2} \\\\ x & = & \dfrac{\pi n}{2}+\dfrac{3\pi}{8} \\\\ x & = & \pi\left( \dfrac{ n}{2}+\dfrac{3 }{8} \right) \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\pi\left( \dfrac{ 4n+3}{8}\right)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\sin(4x)} &\mathbf{=}& \mathbf{\sin(\pi-4x) } = -1 \\\\ \pi-4x &=& \arcsin(-1)+2\pi n,\qquad \mathbf{n \in Z} \\\\ \pi-4x &=& \dfrac{3\pi}{2} + 2\pi n \\\\ 4x &=& 2\pi n-\dfrac{\pi}{2} \quad & | \quad : 4 \\\\ x & = & \dfrac{\pi n}{2}-\dfrac{ \pi}{8} \\\\ x & = & \pi\left( \dfrac{ n}{2}-\dfrac{ 1}{8} \right) \\\\ \mathbf{x} & \mathbf{=} & \mathbf{\pi\left( \dfrac{ 4n-1}{8}\right)} \\ \hline \end{array}\)

Solutions:

\(\begin{array}{|r|c|} \hline & x\text{ on the interval $[0, 2\pi]$} \\ \hline 1 & \dfrac{1}{16}\pi \\ \hline 2 & \dfrac{3}{16}\pi \\ \hline 3 & \dfrac{5}{16}\pi \\ \hline 4 & \dfrac{3}{8}\pi \\ \hline 5 & \dfrac{7}{16}\pi \\ \hline 6 & \dfrac{9}{16}\pi \\ \hline 7 & \dfrac{11}{16}\pi \\ \hline 8 & \dfrac{13}{16}\pi \\ \hline 9 & \dfrac{7}{8}\pi \\ \hline 10 & \dfrac{15}{16}\pi \\ \hline 11 & \dfrac{17}{16}\pi \\ \hline 12 & \dfrac{19}{16}\pi \\ \hline 13 & \dfrac{21}{16}\pi \\ \hline 14 & \dfrac{11}{8}\pi \\ \hline 15 & \dfrac{23}{16}\pi \\ \hline 16 & \dfrac{25}{16}\pi \\ \hline 17 & \dfrac{27}{16}\pi \\ \hline 18 & \dfrac{29}{16}\pi \\ \hline 19 & \dfrac{15}{8}\pi \\ \hline 20 & \dfrac{31}{16}\pi \\ \hline \end{array}\)

How would I solve this equation algebraically?
\(\large{ \sqrt{5x^2+11}} = x + 5\)

\(\begin{array}{|rcll|} \hline \sqrt{5x^2+11} &=& x + 5 \quad & | \quad \text{square both sides} \\ 5x^2+11 &=& (x + 5)^2 \\ 5x^2+11 &=& x^2+10x+25 \\ 4x^2-10x-14 &=& 0 \quad & | \quad : 2 \\ 2x^2-5x-7 &=& 0 \\ \\ x &=& \dfrac{ 5\pm\sqrt{25-4\cdot 2 \cdot (-7) } } {2\cdot 2 } \\\\ x &=& \dfrac{ 5\pm\sqrt{25+56 } } { 4 } \\\\ x &=& \dfrac{ 5\pm\sqrt{81} } { 4 } \\\\ x &=& \dfrac{ 5\pm 9 } { 4 } \\\\ x_1 &=& \dfrac{ 5 + 9 } { 4 } \\\\ x_1 &=& \dfrac{ 14 } { 4 } \\\\ \mathbf{ x_1 } &\mathbf{ =}& \mathbf{ \dfrac{ 7 } { 2 }} \\\\ x_2 &=& \dfrac{ 5 - 9 } { 4 } \\\\ x_2 &=& -\dfrac{ 4 } { 4 } \\\\ \mathbf{ x_2} &\mathbf{ =}& \mathbf{ -1} \\ \hline \end{array}\)

What is the largest integer n such that 3^n is a factor of 1*3*5*...*97*99?

\(\begin{array}{|rcll|} \hline 99!! &=& 1\cdot 3 \cdot 5\cdot 7\cdot \ldots \cdot 95\cdot 97\cdot 99 \\\\ &&\text{Formula:}~\boxed{(2n-1)!! = \frac{(2n)!}{2^n\cdot n!}} \\\\ && (2\cdot 50 - 1)!! = \dfrac{(2\cdot 50)!} {2^{50}\cdot 50! }\\\\ 99!! &=& \dfrac{100!} {2^{50}\cdot 50! } \\ \hline \end{array}\)

100!

The factor \(3^i\)

\(\begin{array}{|rcll|} \hline \frac{100}{3} &=& [33].\overline{3} \\ \frac{100}{3^2} &=& [11].\overline{1} \\ \frac{100}{3^3} &=& [3].\overline{703} \\ \frac{100}{3^4} &=& [1].\overline{234567901} \\ \frac{100}{3^5} &=& 0 \\ \hline 33+11+3+1 &=& 48 \\ i &=& 48 \\ \hline \end{array} \)

50!

The factor \(3^j\)

\(\begin{array}{|rcll|} \hline \frac{50}{3} &=& [16].\overline{6} \\ \frac{50}{3^2} &=& [5].\overline{5} \\ \frac{50}{3^3} &=& [1].\overline{851} \\ \frac{50}{3^4} &=& 0 \\ \hline 16+5+1 &=& 22 \\ j &=& 22 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \dfrac{3^{48}}{3^{22}} \\\\ &=& 3^{48-22} \\ &=& 3^{26} \\ \hline \end{array} \)

\(\text{The largest integer $n$ such that $3^n$ is a factor of $1*3*5*...*97*99$ is $\mathbf{26}$ }\)

compute:

\(\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4}\)

\(\begin{array}{|rcll|} \hline (1+1)^4 &=& \dbinom{4}{0}\cdot 1^4\cdot 1^0 + \dbinom{4}{1}\cdot 1^3\cdot 1^1 + \dbinom{4}{2}\cdot 1^2\cdot 1^2 + \dbinom{4}{3}\cdot 1^1\cdot 1^3 + \dbinom{4}{4}\cdot 1^0\cdot 1^4 \\\\ (1+1)^4 &=& \dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} \\\\ 2^4 &=& \dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} \\\\ 16 &=& \dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} \\ \hline \end{array}\)

Simplify the following expression. Write the answer without negative exponents and use only fractions if necessary (no decimals).

2.)
44c^9x^15y^17/4c^5x^12

\(\begin{array}{|rcll|} \hline &&\dfrac{44c^9x^{15}y^{17}}{4c^5x^{12}} \\\\ &=&\dfrac{44}{4}\cdot\dfrac{c^9x^{15}y^{17}}{c^5x^{12}} \\\\ &=&11\cdot c^9c^{-5}x^{15}x^{-12}y^{17} \\\\ &=&11\cdot c^{9-5}x^{15-12}y^{17} \\\\ &=&11\cdot c^4x^3y^{17} \\ \hline \end{array}\)

Simplify the following expression.
Write the answer without negative exponents and use only fractions if necessary (no decimals).

1.)
24x^2y^7/4x^5y^2

\(\begin{array}{|rcll|} \hline && \dfrac{24x^2y^7}{4x^5y^2} \\\\ &=& \dfrac{24}{4}\cdot\dfrac{x^2y^7}{x^5y^2} \\\\ &=& 6\cdot\dfrac{x^2y^7}{x^5y^2} \\\\ &=& 6\cdot\dfrac{y^7y^{-2}}{x^5x^{-2}} \\\\ &=& 6\cdot \dfrac{y^{7-2}}{x^{5-2}} \\\\ &=& 6\cdot \dfrac{y^5}{x^3} \\ \hline \end{array}\)

Wie spät ist es genau, wenn sich kleiner und großer Zeiger decken, ungefähr um 20 nach vier.

\(\begin{array}{|rcll|} \hline t &=& \dfrac{12}{11}\cdot 4 = 4.\overline{36}^h \Rightarrow \color{red} t = 4^h 21^m49.\overline{09}^s \\ \hline \end{array}\)

b)
Find the coordinates of point N, such that the midpoint of \(\mathbf{\overline{MN}}\) is the origin.

\(M = \left( 3a,-\dfrac{b}{2} \right) \\ N = \left( N_x, N_y \right) \)

1. Solution:

\(\begin{array}{|rclrcl|} \hline \left( \dfrac{3a+N_x}{2}, \dfrac{-\dfrac{b}{2}+N_y}{2} \right) &=& \left( 0, 0 \right) \\\\ \dfrac{3a+N_x}{2} &=& 0 & \dfrac{-\dfrac{b}{2}+N_y}{2} &=& 0 \\ 3a+N_x &=& 0 & -\dfrac{b}{2}+N_y &=& 0 \\ N_x &=& -3a & N_y &=& \dfrac{b}{2} \\\\ \mathbf{N} & \mathbf{=} & \mathbf{ \left( -3a , \dfrac{b}{2} \right) } \\ \hline \end{array}\)

2. Solution:

\(\begin{array}{|rcll|} \hline N &=& -M \\ &=& - \left( 3a,-\dfrac{b}{2} \right) \\ \mathbf{N} & \mathbf{=} & \mathbf{ \left( -3a , \dfrac{b}{2} \right) } \\ \hline \end{array} \)