I think that the sum will always be a rational number
let's prove that
<span>any rational number can be represented as a/b where a and b are integers and b≠0
</span>and an integer is the counting numbers plus their negatives and 0
so like -4,-3,-2,-1,0,1,2,3,4....
<span>so, 2 rational numbers can be represented as
</span>a/b and c/d (where a,b,c,d are all integers and b≠0 and d≠0)
their sum is
a/b+c/d=
ad/bd+bc/bd=
(ad+bc)/bd
1. the numerator and denominator will be integers
2. that the denominator does not equal 0
alright
1.
we started with that they are all integers
ab+bc=?
if we multiply any 2 integers, we get an integer
<span>like 3*4=12 or -3*4=-12 or -3*-4=12, etc.
</span>even 0*4=0, that's an integer
the sum of any 2 integers is an integer
like 4+3=7, 3+(-4)=-1, 3+0=3, etc.
so we have established that the numerator is an integer
now the denominator
that is just a product of 2 integers so it is an integer
<span>2. we originally defined that b≠0 and d≠0 so we're good
</span>therefore, the sum of any 2 rational numbers will always be a rational number <span>is the correct answer.</span>
If the budget is $200 and he have 15 members then we have divide the two. 200 / 15 = $13.33 per shorts. 15x =< $200. x represents 13.33. So the solution represents the coach may spend up to $13.33 per pair of shorts. If it was even 1 cent more than $13.33 than he wouldn't have enough.So he can spend up to $13.33 or less per pair of shorts.
1 is not even yet it is a factor of 8. Alice is wrong.
Answer:
D
Step-by-step explanation:
First, find the coordinates of pointD if A(2,-6), B(10,2) and point D divides line segment AB in the ratio of 5:3.
If point D divides the segment AB in the ratio m:n, then
So
Point C has the x-coordinate the same as point A and the y-coordinate the same as point D.
Thus, C(2,-1)
