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Good morning Synth.

We can make \(4x+3y=3\) into \(y=mx+b\) form! We need to isolate \(y\) .

\(3y=-4x+3, y=-\frac{4}{3}x+1\)

However, the line goes through (2, 3) so the y-intercept may not be 1! Let's substitute.

\(3=-\frac{4}{3}\times 2+b, 3=-\frac{8}{3}+b, b=17/3\) and m=-4/3.

You are very welcome!

:P

Thanks!

You're welcome!

Can you put this in the Off-Topic Posts, please? Thanks 

General Solution - Trigonometry

Sin Ø  - Sin 2Ø = Sin 4Ø - Sin 3Ø

\(\begin{array}{ rcl } \sin(\phi) - \sin(2\phi) &=& \sin(4\phi) - \sin(3\phi) \\ &\text{or} & \\ \sin(3\phi)+\sin(\phi) &=& \sin(4\phi) + \sin(2\phi) \\ \end{array} \)

Formula:

\(\begin{array}{|rcll|} \hline \underbrace{\sin(3\phi)+\sin(\phi)}_{=2\sin(2\phi)\cos(\phi)} &=& \underbrace{\sin(4\phi) + \sin(2\phi)}_{=2\sin(3\phi)\cos(\phi)} \\\\ 2\sin(2\phi)\cos(\phi) &=& 2\sin(3\phi)\cos(\phi) \\\\ \sin(2\phi)\cos(\phi) &=& \sin(3\phi)\cos(\phi) \\\\ \sin(3\phi)\cos(\phi) -\sin(2\phi)\cos(\phi) &=& 0 \\\\ \cos(\phi)\Big(\sin(3\phi)-\sin(2\phi) \Big) &=& 0 \\ \hline \end{array} \)

Formula:

\(\begin{array}{|rcll|} \hline \cos(\phi)\Big(\underbrace{\sin(3\phi)-\sin(2\phi)}_{=2\cos(\frac52\phi)\sin(\frac12\phi)} \Big) &=& 0 \\\\ \cos(\phi)2\cos(\frac52\phi)\sin(\frac12\phi) &=& 0 \quad | \quad :2 \\\\ \large{\mathbf{\cos(\phi)\cos(\frac52\phi)\sin(\frac12\phi)} }& \large{\mathbf{=}} & \large{\mathbf{0}} \\ \hline \end{array}\)

General Solution:

\(\begin{array}{|lrcll|} \hline 1. & \cos(\phi) &=& 0 \\ & \phi &=& 2n\pi\pm\arccos(0) \\ & \phi &=& 2n\pi\pm \frac{\pi}{2} \quad \Rightarrow \quad \mathbf{\phi = (2n-1)\frac{\pi}{2}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 2. & \cos(\frac52\phi) &=& 0 \\ & \frac52\phi &=& 2n\pi\pm\arccos(0) \\ & \frac52\phi &=& 2n\pi\pm \frac{\pi}{2}\quad | \quad \cdot \frac25 \\ & \phi &=& \frac45 n\pi\pm \frac{\pi}{5} \quad \Rightarrow \quad \mathbf{\phi = (2n-1)\frac{\pi}{5}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3. & \sin(\frac12\phi) &=& 0 \\ & \frac12\phi &=& n\pi+(-1)^n \arcsin(0) \\ & \frac12\phi &=& n\pi+(-1)^n \pi \quad | \quad \cdot 2 \\ & \phi &=& 2n\pi+(-1)^n 2\pi \quad \Rightarrow \quad \mathbf{\phi = 2n\pi} \\ \hline \end{array} \)

14500(4/5)^n

Slope is 4

y=4x+b      includes point (3,1)

1 = 4(3) + b

-11 = b

y = 4x -11

Rewrite   4x+3y=3

               3y=-4x+3

                 y = -4/3x +1      where =-4/3 is the slope....parallel line will have same slope

y = -4/3 x + b    includes 2,3

3 = -4/3 (2) + b

3+8/3 = b

b = 5 2/3

so NEW line is    y = -4/3x + 5 2/3      m = -4/3   b = 5  2/3

Which one is abc form ?

L(I)=10log⁡(I/10⁻¹²)

L(I) = 125, so  125 = 10log⁡(I/10⁻¹²)

Divide both sides by 10

12.5 = log⁡(I/10⁻¹²)

Antilog both sides:

10^12.5 = I/10⁻¹²

Multiply both sides by 10^-12 

10^-1210^12.5 = I

Using the fact that 10^a10^b = 10^a+b we have:

I = 10^0.5   or  I = √10

Adam has four short straws, five medium straws, and six long straws.
If he randomly draws two straws, one at a time without replacement,
what is the probability that both are medium straws?

\(\dfrac{5}{15}\times \dfrac{4}{14} = 0.09523809524\quad (9.5\ \%)\)

There are 32 students in the Classroom. There 12 boys and 20 girls.

She needs to select two students to represent her class .

She will put all students names on a piece of paper and randomly select two names, one after the other.

What is the probability that she will select a boy first and then a girl?

\(\dfrac{12}{32}\times \dfrac{20}{31} = 0.24193548387\quad (24.2\ \%) \)

buy low sell high

Thank you, CPhill

Hello CPhill,

"how was that method derived":

1.)

This is the Legendre Theorem from 1808:

Yes I forgot the 0.

No big deal, I showed you how to do it. 

You can pick up trivial errors like that on your own.

You do not need to be sorry - I did it correctly. 

A thank you would be nice though.