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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.