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>
First simplify the number part, 14/2 = 7/1
then the x part,
y doesn't need to be simplified
then simplify the z part, z^-6 = z^6
put it all together...
Step 1: All angles in a triangle must add up to 180! Therefore, in the upper triangle, add up the two you know 85+35=120 so the missing angle is 60.
Step 2: Bottom triangle. Your new angle above (60) is the pair of the opposite angle - so it too is 60. So now you know 2 of the 3 angles: 60+64=124 Subtract that from 180 and you get 56...which is the answer for ?
Answer:
-11(j+3k)
Step-by-step explanation:
-11j-33k
We can factor out a negative 11 from each term
-11j = -11 *j
-33k = -11 *3k
-11j -33k = -11(j+3k)
