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>
Answer:
-16b+40b^2
Step-by-step explanation:
-8b*2=-16b
(-8b)(-5b)=40b^2
Answer:
The answer is 60 and 80
Step-by-step explanation:
We know that William
paid his deductible that amount is the $150.
Then William paid 20 percent of everything else. That's 20 percent of (750
minus 150), it is equal to 150.
Then the insurance company paid 80 percent of (750 minus 250), it is equal to
480.
