Solveit

how many black card s are there ?

Answer is (A) you wrote it wrong

the answer should be like that if half of the card s are black :

A.1/2,25/51, 12/25, 23/49

put in term s of a any number

:D                  

Thanks ElectricPavlov !

\(\frac{i}{i+1}=\frac {i(i-1)}{(i+1)(i-1)}=\frac {-1-i}{-1-1}=\frac{1+i}{2}=\frac12+\frac 12i\)

as the same as \(a+bi\)

\(x^2-10x=0\\ x(x-10)=0\\ x_1=0\\ x_2=10\)

1+1=2

\(553\div5=110.6\)

 there is 110 numbers wich is devisible by 5

\(553\div3=184.333\)

there is 184 numbers wich is devisible by 3

\(553\div15=36.9\)

there is 36 numbers wich is devisible by both 5 and 3

now let s find sum of devisors of 5:

\(5+5\times2+5\times3+5\times4\dots+5\times110\\ 5(1+2+3+4\dots+110)\\ 5(55.5\times110)=30\ 525\)

sum of devisors of 3:

\(3(92.5\times184)=51\ 060\)

sum of devisors of both 5 and 3

\(15(18.5\times36)=9990\)

\(30\ 525+51\ 060-9990=71\ 595\)

\(-1+\frac1{2\times5}=-1+\frac{1}{10}=\frac{-10+1}{10}=-\frac{9}{10}\)

\(1+\frac12=\frac{2+1}{2}=\frac32\)

\(\begin{array}{rcll}2^3 \times 2^x &=& 2^{3x}\\ 2^{3+x}&=&2^{3x}\\ 3+x&=&3x\\ 3&=&2x\\ 1.5 &=&x \end{array}\)