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>
(1/2)x+b=y because the equation that given the slope is 1/2 if you leave y alone and take the x’s number
put the point now , 4+b=3 so b is -1
the equation is (1/2)x-1=y
Answer:
x ≈ 1.4
Step-by-step explanation:
Using the sine ratio in the right triangle
sin50° = 
= 
= 
( multiply both sides by 1.8 )
1.8 × sin50° = x , then
x ≈ 1.4 ( to the nearest tenth )
Answer:
x = - 2
y = 2
Step-by-step explanation:
y = - 2 + 4
y = 2
- 2 + 2(2) = 2
- 2 + 4 = 2
