Given 11 and 6, which of the following could not be the third side of the triangle?

      Any, that is greater than 5 and less than 17

Step-by-step explanation:

Given three sides of triangle: a, b and c, and a<c and b<c, then the triangle exist only if  a+b>c

therefore:

if a = 11 and b = 6 and c = ??

then:  11 + 6 > c

            c < 17

if a = 6 and c = 11 and b = ?? {or b=6 and a=??}

then:  6 + b > 11  {subtract 6 from both sides}

            b > 5

To write y-3=2/3(x-2), you must change the equation to this format: ax+by=c, whereas there are no common factors between the coefficients and constant, no fractions or decimals, and the coefficient of a cannot be negative. To start, first multiple both sides by 3 to get rid of the fraction 2/3. This would result in this: 3y-9=2(x-2)
Then, you distribute the 2 to (x-2) and you get 3y-9=2x-4
Now, to get this equation into ax+by=c form, subtract 2x from both sides, and add 9 to both sides, which will result in this: -2x+3y=5.
Now, since the coefficient of a is negative, multiply the entire equation by -1, to get you 2x-3y=-5