Take any arbitrary integer n and then square it, which one is true? n2 cannot be in the form 7k + 5. O n? cannot be in the form 7k + 4. n² cannot be in the form 7k + 1. O n? cannot be in the form 7k. O n? cannot be in the form 7k + 2.

Using c++

Click and drag the steps in the correct order to show that if n is a positive integer, then n is even if and only if 7n +4 is even. Suppose that n is even. Suppose that n is not even. Thus, 7n + 4 = 14k +10= 2(7k+5). Then, 7n + 4 = 14k + 2 = 2(7k + 1). Since n is even, it can be written as 2k for some integer k. This is 2 times an integer; so it is even as desired. Then, n can be written as 2k +1 for some integer k. Thus, 7n + 4 = 14k +11 = 2(7k+5)+1. Then, 7n + 4 = 14k + 4 = 2(7k+ 2). This is 1 more than 2 times an integer, so it is odd. Reset

7. Find m/CAB and mZABC. 8. Find m/CAB and mZABC. 126* (5x+10) C A (8x-4)* (4x+4)* (4x-1) B B