Given:
x, y and z are integers.
To prove:
If 
is even, then at least one of x, y or z is even.
Solution:
We know that,
Product of two odd integers is always odd. ...(i)
Difference of two odd integers is always even. ...(ii)
Sum of an even integer and an odd integer is odd. ...(iii)
Let as assume x, y and z all are odd, then is even.
is always odd. [Using (i)]
is always odd. [Using (i)]
is always even. [Using (ii)]
is always odd. [Using (iii)]
is always odd.
So, out assumption is incorrect.
Thus, at least one of x, y or z is even.
Hence proved.
Answer:
c
Step-by-step explanation:
because a function has no repeating x values. and 5 repeats
Answer:
1. (2/3, -2)
2. (-3/2, 3)
3. (9/4, 3)
4. (4/5, -1)
5. (doesn't show the question, ill answer if u tell me what it is)
6. (4/3, 4)
7. (-3, 3/2)
8. (-2, 3)
9. (3, -2)
10. (doesn't show the question, ill answer if u tell me what it is)
Step-by-step explanation:
I'm a master at graphing :3. Also just if you forgot or don't know it's like (x, y). Hope I helped :)
Answer:
infinite solutions
Step-by-step explanation:
Hello!
Since 3r - 5 is on both sides of the equation we can put any number in for r and it would still be true so the answer is infinite solutions
The answer is infinite solutions
Hope this helps!
Step-by-step explanation:
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