So we can start with the full of possibilities and eliminate them one by one.
The full set is {0,1,2,3,4,5,6,7,8,9}.
Now we know that any prime greater than 2 is odd as otherwise it would have 2 as a factor, so we can eliminate all of these digits that would be an even number, leaving:
{1,3,5,7,9}
We also know that any prime greater than 5 cannot be a multiple of 5 and that all numbers with 5 in the digits are a multiple of 5, so we can eliminate 5.
{1,3,7,9}
We know that 11,13,17 and 19 are all primes, so we cannot eliminate any more of these, leaving the set:
Given a quadratic ax² + bx + c, find factors of ac that add up to b. Divide those factors by a and reduce. The denominators become the coefficients and the numerators become the constants.
Step-by-step explanation: First, you need to draw a picture and label the parts of the line: AB=5x-15; BC= 3x-5; AC =28. Because of the segment addition postulate, you set the equation to be 5x-15+3x-5=28. Then you solve:
5x-15+3x-5=28
Add like terms:
8x-20=28
Add 20 to both sides
8x=48
Divide by 8
x=6
Now, you need to find the measure of AB, so you plug the 6 into the x variable for 5x-15