104
(3r + 2)(r - 5)
285
1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 = 1 + 1 + 1 + 1 + 1 + 1 = 5
It is: 36 plus 121 = 157
104
No
(3r + 2)(r - 5)
(5r - 4)(5r + 4)
domain = set R ,, all real numbers
6 squared plus 9 squared is equal to 117.
12 squared plus 18 squared is equal to 468.
25 squared plus 49 squared is equal to 3,026.
It could be lots of things. One answer can be: r2 + 9 = r2 + 32.
1 squared plus 8 squared or 4 squared plus 7 squared
(a + x^2)(b + y^2)
(r+5)(r-4) The idea is to get two numbers whose product in this case is -20, and whose sum in this case s +1.