Did you forget to finish the question
Answer:
C. Yes Devon should have evaluated f(3).
Step-by-step explanation:
She substituted the wrong value. For x - 3 to be a factor f(3) must equal 0.
f(3) = 3^3 + 3^2 - 10(3) + 8 = 27 + 9 - 30 + 8
= 14, so x- 3 is not a factor anyway but she did substitute the wrong value.
Answer:
No, mn is not even if m and n are odd.
If m and n are odd, then mn is odd as well.
==================================================
Proof:
If m is odd, then it is in the form m = 2p+1, where p is some integer.
So if p = 0, then m = 1. If p = 1, then m = 3, and so on.
Similarly, if n is odd then n = 2q+1 for some integer q.
Multiply out m and n using the distribution rule
m*n = (2p+1)*(2q+1)
m*n = 2p(2q+1) + 1(2q+1)
m*n = 4pq+2p+2q+1
m*n = 2( 2pq+p+q) + 1
m*n = 2r + 1
note how I replaced the "2pq+p+q" portion with r. So I let r = 2pq+p+q, which is an integer.
The result 2r+1 is some other odd number as it fits the form 2*(integer)+1
Therefore, multiplying any two odd numbers will result in some other odd number.
------------------------
Examples:
- 3*5 = 15
- 7*9 = 63
- 11*15 = 165
- 9*3 = 27
So there is no way to have m*n be even if both m and n are odd.
The general rules are as follows
- odd * odd = odd
- even * odd = even
- even * even = even
The proof of the other two cases would follow a similar line of reasoning as shown above.
Answer:
V=56
Step-by-step explanation:
7x8=56
56/7=8
