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The total number of ways that the numbers can be chosen is 40c4. Exactly 10 of these possibilities result in the four slips having the same number.

Now we need to determine the number of ways that two slips can have a number  a and the other two slips have a number b , with \($b\ne a$\) . There are  10c2 ways to choose the distinct numbers a and b . For each value of a , there are 10c2 ways to choose the two slips with a  and  for each value of b , there are 4C2 ways to choose the two slips with b . Hence the number of ways that two slips have some number  and the other two slips have some distinct numberb  is

(10C2)(4c2)(4c2)=45*6*6=1620

So the probabilities  p and q  are 10/40c4 and 1620/40c4 , respectively, which implies that

q/p=1620/10=162

To find the number of positive integers that satisfy both conditions, we can consider the possible combinations of digits (1, 3, 5) that add up to 6.

Let's break down the possible cases:

Three 1's: 1+1+1+1+1+1=6

One 3 and three 1's: 3+1+1+1=6 

One 5 and one 1: 5+1=6

So, there are three possible combinations that satisfy both conditions. Therefore, there are three positive integers: 111111, 3111,1311,1131,1113.15 and 51.

(a+2)(3a^2-2a-12)(a^2-2a+1)

=(3a^3-2a^2-12a+6a^2-4a-24)(a^2-2a+1)=(3a^3+4a^2-16a-24)(a^2-2a+1)=3a^5+4a^4-16a^3-24a^2-6a^4-8a^3+32a^2+48a+3a^3+4a^2-16a-24

=3a^5-2a^4-21a^3+12a^2+32a-24

The quadratic roots are 1 and -1.then, the factor of the quations are x-1 and x+1.

SO, our answer is (x-1)(x+1)=x^2-1

BY triangles iniquality 

TU+UV>TVTHEN,6x-11+3x-1>2x+3 then calculating this we got x>15/7

UV+TV>TUthen,3x-1+2x+3>6x-11,then claculating this we got 13>x.

TV+TU>UV then,2x+3+6x-11>3x-1then calculating this we got x>7/5.

ANSWER IS 13>X>15/7.(THESE ARE THE POSSIBLE VALUES OF X.)

WE GOT X WHEN X MULTIPLY BY CONSTANT OR CONSTANT MULTIPLY BY X TERMS . 

HERE ,WHEN X IS MULTIPLY BY 14 THEN WE GOT 14X.

ALSO -23X IS MULTIPLY BY 1 WE GOT -23X.

-23X+14X=-7X.

SO , OUR ANSWER IS -7.

Since we defined  g(x)=f(x/2), a real number x is in the domain of g if and only if x/2  is in the domain of f . Therefore, the domain of g consists of all x such that

-8<=x<=8

The solutions to this inequality are given by -16<=x<=16.

so , the domain of g is interval of width 16-(-16)=32.

ANSWER IS 33.

SOLVING THE EQ (2)\(\Rightarrow\)

     2(X+4)+2Y=11+1/2(16+2X)

\(\Rightarrow\)2X+8+2Y=11+8+X

\(\Rightarrow\)X+2Y=11 

\(\Rightarrow\)3X+6Y=33.

SOLVING THE EQ(1)\(\Rightarrow\)

    X+6Y=R-2X

\(\Rightarrow\)3X+6Y=R.

THE EQ ARE INFINITE SOLULTION WHEN 

a1/a2=b1/b2=c1/c2

HERE

3X/3X=6Y/6Y=33/R=1

SOLVING FOR R THEN, R=33.

SORRY ,BUT YOUR ANSWER IS WRONG.