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Solve:

\(|2x - 1| = |3x + 5|\).

Write your answers as a list of numbers, separated by commas,
e.g. '23, 19' (but without the quotes).

\(\begin{array}{|rcll|} \hline |2x - 1| &=& |3x + 5| \\ (2x - 1)^2 &=& (3x + 5)^2 \\ 4x^2-4x+1&=& 9x^2+30x+25 \\ \ldots \\ 5x^2 +34x+24 &=& 0 \\\\ x &=& \dfrac{-34\pm \sqrt{34^2-4\cdot 5 \cdot 24} }{2\cdot 5} \\ &=& \dfrac{-34\pm \sqrt{34^2-480} }{10} \\ &=& \dfrac{-34\pm \sqrt{676} }{10} \\ &=& \dfrac{-34\pm 26 }{10} \\\\ x &=& \dfrac{-34+ 26 }{10} \\ \mathbf{x} & \mathbf{=} & \mathbf{-0.8}\\\\ x &=& \dfrac{-34- 26 }{10} \\ \mathbf{x} & \mathbf{=} & \mathbf{-6}\\ \hline \end{array} \)

\(\mathbf{"-0.8, -6"}\)

Maybe this question is easier than it looks.

One train is traveling at 12km/hr and the other is traveling at 20km/hour in the opposite direction.

If you are a passenger sitting inside the train it appears like you are stopped and everything outside the train is moving.

The train you are NOT on has the relative speed of 12+20=32km/hour

So I think the senario is similar to  a magical soft brick wall and a train that is at an instant 72 km away traveling towards it at 32km/hour.

s=d/t

we have distance and speed so t = d/s = 72/32 = 2.25 hours.

The two trains will collide very gently in 2.25 hours.

All this time poor crazy hyde is flying.  So he flies at 48km/hour for 2.25 hours

distance he has travelled is  s*t = 48*2.25 = 108km.

For 1. note that \(|z| = |a+ib|=\sqrt{a^2+b^2}\)

For 2. 

For a certain value of k , the system

x + y + 3z = 10
-4x + 2y + 5z = 7

kx + z = 3
has no solutions. What is this value of k ?

For a certain value of k , the system
\(x + y + 3z = 10 \\ -4x + 2y + 5z = 7 \\ kx + z = 3 \)
has no solutions.

What is this value of k ?

\(\begin{array}{|rcll|} \hline \begin{vmatrix} 1 & 1 & 3 \\ -4 & 2 & 5 \\ k & 0 & 1 \\ \end{vmatrix} &=& 0 \\\\ 1\cdot \begin{vmatrix}2&5\\ 0&1 \\\end{vmatrix} -1\cdot\begin{vmatrix}-4&5\\ k&1 \\\end{vmatrix} +3\cdot\begin{vmatrix}-4&2\\ k&0 \\\end{vmatrix} &=& 0 \\\\ 2-(-4-5k)+3(-2k) &=& 0 \\ 2+4+5k-6k &=& 0 \\ 6-k &=& 0 \\ \mathbf{k} & \mathbf{=}& \mathbf{6} \\ \hline \end{array} \)

\(\text{the first number must be odd}\\ 4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 144\\\)

1) we need to know what the radius of the circle is

2) 7 laps is 21 miles, 10 min is 1/6 hr.  (21miles)/(1/6 hr) = 126 miles/hr

y^6 is the answer

1. 360 degrees/60 degrees = 6

The circumference of the circle is 20pi

20pi/6 = 10pi/3

so C

2.360 degrees/60 degrees = 6

The area of the circle is 100pi

100pi/6 = 50pi/3

so A

Well BC is easy to find.  Use the area of a triangle formula and rearrange it. 

then you have the base and the height of ABD so you can just use the area of a trianlge to get the area.

Thanks EP  

https://web2.0calc.com/questions/help-plz-sigma-notations-a-bit-hard-for-me

Pick a number for the smaller circle diameter.....say 2 units....the larger circle diameter is 4

Area of circle = pi r^2

Smaller/Larger  =   pi 1^2  / pi 2^2  =  1^2/2^2 =   1/4     or   1:4

10 is not prime.

I just got the site calculator to give me the factors.  I didn't check.

3*5*7*2*2*2 = 840    

Woops it was 8400 not 840    

Ok I am glad you caught my copy error!