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:
A
Step-by-step explanation:
1:2.5 because 2/2=1 and 5/2=2.5
Mischa wrote the quadratic equation 0 = –x2 + 4x – 7 in standard form
The standard form of quadratic equation is
When y=0 then the standard equation becomes
Now we compare the given equation 
with and find the value of a, b,c
-x^2 can be written as -1x^2
a= -1 , b=4 and c=-7
The value of c is -7
Answer: 2
Step-by-step explanation:
theres a chance i could be right
This is hard to answer without a picture but with this I would say 33% ?
