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: Option 'B' and 'E' are correct.
Step-by-step explanation:
Since we have given that
Simplification needs a few steps:
So, Option 'B' and 'E' are correct.
As Option 'B' says there are 3 terms in the simplified product, which is true.
14 times
To evaluate this, you would divide 84 by 6, which is equal to 14.
Answer:
5,8 or 8,5
Step-by-step explanation:
A arcade game gives 8 tickets this = y
For every 5 game played this = x
so we can make the assumption that every 5 games played he gets 8 tickets you can write this as 5,8 or. Every 5 game played is 8 tickets.
15.50 x 20 = $310
31 x 8 = $248
310 + 248 = $558
Her total pay for the week is $558.
Hope you got a good grade!!! :)
