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Guest is correct, but next time I can cross out the words and such.

Here is a hastily written one line code in C++ that does the same thing in 1 nano second:

a=123; b=int(log(a));c=int(a / 10^b);d=int(a / 10^(b-1)) %10; e=a%10; f=e*(10^b) + d*(10^(b-1)) + c*(10^(b-2))=321

I am not actually sure what you are being asked to do.

I would assume it means  R2 is replaced by R2- 1/2 R1

in which case the answer would be

\(\begin{bmatrix} 4 && 3 && 0 \\ -6-2 && -3-1.5 && 12-0 \end{bmatrix}=> \begin{bmatrix} 4 && 3 && 0 \\ -8 && -4.5 && 12 \end{bmatrix}\)

Thank you, Melody !

That is really cool.   Thanks Heureka.

https://study.com/academy/lesson/what-is-scientific-notation-definition-rules-examples.html

https://www.google.com/search?q=scientific+notation+parts&tbm=isch&source=univ&client=firefox-b-1-d&sa=X&ved=2ahUKEwjj0cOPzuPkAhWFsJ4KHRztDKAQsAR6BAgHEAE&biw=1920&bih=916

https://www.yourdictionary.com/mantissa

Basically, the situation here to be understood is that an angle inscribed (vertex on the circle perimeter) intercepts an arc ofTWICE the degrees of the angle .

#9  16x-10 = 2 (67)   solve for x       then put x back into the 16x-10 to find the arc degrees

#10  2 (7x+9) = 46

#11 2 (15x+2) = 360 - 87-39

#12  360 - ( 75 + 2 (59)) = 17x-20     

  Cool? 

Basically, the situation here to be understood is that an angle inscribed (vertex on the circle perimeter) intercepts an arc ofTWICE the degrees of the angle .

#9  16x-10 = 2 (67)   solve for x       then put x back into the 16x-10 to find the arc degrees

#10  2 (7x+9) = 46

#11 2 (15x+2) = 360 - 87-39

#12  360 - ( 75 + 2 (59)) = 17x-20     

  Cool? 

Eventually....at some pont, the captcha robot thingie quits happening....I do not know what the tripping point is.... ~ EP

Why 5-x marbles? Is it because that guaruntees martin will choose 4, 3, 2, or 1 marbles?

Hello CalculatorUser

No, it is because that guaruntees  that after Martin and Ron the sum of removed marbles is 5.

or in other words Ron supplements each time to 5.

Here is the course

\(\begin{array}{|r|r|l|} \hline \text{Martin} & \text{Ron} & \text{sum} \\ \hline & 3 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ 1 \\ \hline \end{array}\)

I think he can copy and paste it. 

Why 5-x marbles? Is it because that guaruntees martin will choose 4, 3, 2, or 1 marbles?

Ron and Martin are playing a game with a bowl containing 39 marbles.
Each player takes turns removing 1, 2, 3 or 4 marbles from the bowl.
The person who removes the last marble loses.
If Ron takes the first turn to start the game,
how many marbles should he remove to guarantee he is the winner?

Ron first takes 3 marbles.

Than, if Martin takes x marbles, Ron takes 5-x marbles, this repeats 7 times.

The last marble is then for Martin.

39 = 3 + 7*5 + 1

Yes agreed!! thats why Im confused... The answer was apparently 3 but I don't know why.

No matter how many marbles he takes with his first choice Ron can still lose if he is not playing strategically.

I guess we are suppose to assume that Ron is playing the ultimate game and so it Martin.

I have got nothing more.  

2x + 8x² + 9 - (4 - 2x - 8x²)

There is a negative 1 before the parantheses

2x + 8x² + 9 - 4 + 2x + 8x²

Which simplifies to

16x² + 4x + 5

Good job Power27!
 

How were able to decipher the picture code?

Coolstuff,

Is CalculatorUser supposed to be able to find your edits?

Ignore the 57 degree angles

Angle W and the 87 degree angle add up to 180

180 - 87 = 93

Angle W is 93 degrees

Angle V and W are congruent

Angle V is 93 degrees

Angle V and X are supplementary

So angle x is 180 - 93 = 87 degrees

Thanks Chris,

Another one for me to study   

Ok I have studied it. 

I need to work on similar figure geometry questions too,  (Along with CalculatorUser.)

In summary you have said

Area desired = area of BPQ - area of BNM

These are similar triangles.

To find these areas you need length MN and length PQ and PB

You have then used a series of similar figures ratios and pythagoras theorum to do this.

All sorted.    

Triangle BPC is an isossilies triangle because sides BP and BC are the same length, so angles BPC and BCP are congruent. To find angle BPC you first need to find angle PBC.

Triangle APB is equilateral so all the angles measure 60 degrees, so angle APB measures 60 degrees. All the angles in a square measure 90 degrees. The measure of angle PBC is the sum of the angles ABC=90 and ABP=60 so PBC=150 degrees.

\(2 \cdot m\angle BPC+m\angle PBC=180\)

150 can be substituted for angle PBC

\(2 \cdot m\angle BPC+150=180\)

subtract 150

\(2 \cdot m\angle BPC=30\)

divide by 2

\(\boxed{m\angle BPC=15}\)

Good job, Power27!