heureka

Matrix Rotation counterclockwise:

\(\begin{array}{|rcll|} \hline \begin{pmatrix} \cos(\varphi) & \sin (\varphi) \\ -\sin(\varphi) & \cos (\varphi) \\ \end{pmatrix} \stackrel{\varphi=270^{\circ}} \rightarrow \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} \\ \hline \end{array} \)

The point P becomes to P':

\(\begin{array}{|rcll|} \hline \binom{x}{y}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{y}{-x} \\ \hline \end{array} \)

\(\text{Let}\ X =\binom{-2}{1} \\ \text{Let}\ Y =\binom{-4}{2} \\ \text{Let}\ Z =\binom{-3}{5} \\ \)

\(\begin{array}{|rcll|} \hline X'=\binom{-2}{1}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{1}{2} \\ Y'=\binom{-4}{2}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{2}{4} \\ Z'=\binom{-3}{5}\cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} = \binom{5}{3} \\ \hline \end{array} \)

The answer is c.

Find the 5th term of the expansion of the binomial (3x-1)⁷

\(\begin{array}{|rcll|} \hline (3x-1)^7 &=& \binom{7}{0}\cdot (3x)^7 + \binom{7}{1}\cdot (3x)^6\cdot(-1)^1 \\ &+& \binom{7}{2}\cdot (3x)^5\cdot(-1)^2 + \binom{7}{3}\cdot (3x)^4\cdot(-1)^3 \\ &+& \color{Maroon}\binom{7}{4}\cdot (3x)^3\cdot(-1)^4\color{black} + \binom{7}{5}\cdot (3x)^2\cdot(-1)^5 \\ &+& \binom{7}{6}\cdot (3x)^1\cdot(-1)^6 + \binom{7}{7}(-1)^7 \\ \hline \end{array} \)

The 5th term of the binomial (3x-1)⁷  is:

\(\begin{array}{|rcll|} \hline && \color{Maroon}\binom{7}{4}\cdot (3x)^3\cdot(-1)^4\color{black} \\ &=& \binom{7}{4}\cdot (3x)^3 \quad & | \quad \binom{7}{4} = \binom{7}{7-4} = \binom{7}{3} \\ &=& \binom{7}{3}\cdot (3x)^3 \\ &=& \binom{7}{3}\cdot 27x^3 \\ &=& \frac{7}{3}\cdot\frac{6}{2}\cdot\frac{5}{1}\cdot 27x^3 \\ &=& 7\cdot\frac{6}{6}\cdot 5\cdot 27x^3 \\ &=& 35\cdot 27x^3 \\ &\mathbf{=}& \mathbf{945x^3} \\ \hline \end{array}\)

is 27, 36, 45 a pythagorean triad

\(\begin{array}{|rcll|} \hline \sqrt{27^2+ 36^2} = 45 \qquad \text{yes 27, 36, 45 a pythagorean triad} \\ \hline \end{array} \)

What's the derivative of sqrt(3)x^5-6x^4+54*sqrt(7)*x^3+2x^2-5/7x+34.16?

\(\begin{array}{rcll} && \dfrac{d}{dx}\left(\sqrt{3}\ x^5 - 6 x^4 + 54 \sqrt{7}\ x^3 + 2 x^2 - \frac{5 x}{7} + 34.16 \right) \\\\ &=& 5 \sqrt{3}\ x^4 - 24 x^3 + 162 \sqrt{7}\ x^2 + 4 x - \frac57 \\ \end{array}\)

sin(a+b)=0.86

sin(a-b)=0.34

What is sinbcosa?

\(\small{ \begin{array}{|lrcll|} \hline (1) & \sin(a+b) &=& \sin(a)\cos(b)+\cos(a)\sin(b) \\ (2) & \sin(a-b) &=& \sin(a)\cos(b)-\cos(a)\sin(b) \\ \hline (1)-(2): & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)+\cos(a)\sin(b)-[~\sin(a)\cos(b)-\cos(a)\sin(b)~] \\ & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)+\cos(a)\sin(b)- \sin(a)\cos(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& \sin(a)\cos(b)- \sin(a)\cos(b)+\cos(a)\sin(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& \cos(a)\sin(b)+\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& 2\cos(a)\sin(b) \\ & \sin(a+b)-\sin(a-b) &=& 2\sin(b)\cos(a) \\ & \sin(b)\cos(a) &=& \frac12\cdot [\sin(a+b)-\sin(a-b)] \quad | \quad \sin(a+b)=0.86 \quad \sin(a-b)=0.34 \\ & \sin(b)\cos(a) &=& \frac12\cdot (0.86-0.34) \\ & \sin(b)\cos(a) &=& \frac12\cdot (0.52) \\ & \mathbf{\sin(b)\cos(a)} & \mathbf{=} & \mathbf{0.26} \\ \hline \end{array} } \)

I gave a shot and I'm not sure if I got it correct by sheer luck or if my methods were correct.

Your methods are correct.

I ended up having to use a calculator at the (-8)^3 part as I wasn't sure how to solve for that in any other way than brute forcing it.

\(\begin{array}{|rcll|} \hline &&(-8)^3 \pmod {27} \\ &\equiv& (-8)^2\cdot(-8) \pmod {27} \\ &\equiv& 64\cdot(-8) \pmod {27} \quad & | \quad 64 \pmod {27} \equiv 64-2\cdot 27 \pmod {27} = 10\pmod {27} \\ &\equiv& 10\cdot(-8) \pmod {27} \\ &\equiv& -80 \pmod {27} \\ &\equiv& -80 + 3\cdot 27 \pmod {27} \\ &\equiv& -80 + 81 \pmod {27} \\ &\equiv& 1 \pmod {27} \\ \hline \end{array}\)

10^123 = 2^x

find x

\(\begin{array}{|rcll|} \hline 10^{123} &=& 2^x \quad & | \quad \log_{10}() \\ \log_{10}(10^{123}) &=& \log_{10}(2^x) \\ 123\cdot \log_{10}(10) &=& x\cdot \log_{10}(2) \quad & | \quad \log_{10}(10) = 1 \\ 123 &=& x\cdot \log_{10}(2) \\ x\cdot \log_{10}(2) &=& 123 \\ x &=& \frac{123}{\log_{10}(2)} \\ x &=& \frac{123}{0.30102999566} \\ \mathbf{x} & \mathbf{=} & \mathbf{408.597155671} \\ \hline \end{array}\)

Hi Melody,

i have installed Latex on my Computer. The setup include a TexStudio.

See picture:

all free-software.

Then i have snapped it as image into web2.0calc.

I do it all directly on my hard drive.

Here i think we have a minimal latex.

For what values of a does the system of equations of linear equations
in the variables x and y have infinitely many solutions?
\(\begin{array}{|rcll|} \hline ay &=& -\frac{3}{4}x \\ 4x-5y &=& 0 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline ay &=& -\frac{3}{4}x \quad &| \quad + \frac{3}{4}x \\ \frac{3}{4}x + ay &=& 0 \quad &| \quad \cdot \frac{16}{3} \\ 4x + \frac{16}{3}ay &=& 0 \\ \hline \end{array}\)

infinitely many solutions, if  \(\frac{16}{3}a = -5\)

\(\begin{array}{|rcll|} \hline \frac{16}{3}a &=& -5 \quad &| \quad \cdot \frac{3}{16} \\ a &=& -5\cdot \frac{3}{16} \\ a &=& -\frac{15}{16} \\ \hline \end{array}\)