MHB Is the Calculation of $(3 \diamond 8) \square 2 = 1940$ Correct?

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Binary Composite
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
define the binary operater $\diamond$ by
$a\diamond b$=4a+4b
and $s\square b$ by $a\square b=a^2+b^2$
find
$(3\diamond 8)\square 2=
[(4(3)+4(8))^2 +2^2]=1940$
ok just want to see if this is correct
before i run thru the ribbon :unsure:
 
Last edited:
Mathematics news on Phys.org
This is correct.
 
1940
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top