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:
x=-1
Step-by-step explanation:
Answer:
1668.75
Step-by-step explanation:
Answer:
77. 
Proved
78. ![\sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ]](https://tex.z-dn.net/?f=%5Csec%5E%7B4%7Dx%20%5Ctan%5E%7B2%7D%20x%20%3D%20%5Csec%5E%7B2%7Dx%20%5B%5Ctan%5E%7B2%7Dx%20%2B%20%5Ctan%5E%7B4%7Dx%20%5D)
Proved
79. ![\cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x](https://tex.z-dn.net/?f=%5Ccos%5E%7B3%7D%20x%5Csin%5E%7B2%7D%20x%20%3D%20%5B%5Csin%5E%7B2%7Dx%20-%20%5Csin%5E%7B4%7Dx%5D%20%5Ccos%20x)
Proved.
80. 
Proved.
Step-by-step explanation:
77. Left hand side
= 
= 
= ![\cot^{4}x [\csc^{2}x - 1]](https://tex.z-dn.net/?f=%5Ccot%5E%7B4%7Dx%20%5B%5Ccsc%5E%7B2%7Dx%20-%201%5D)
{Since we know, 
}
= 
= Right hand side (Proved)
78. Left hand side
= 
= ![\sec^{2} x [1 + \tan^{2}x] \tan^{2} x](https://tex.z-dn.net/?f=%5Csec%5E%7B2%7D%20x%20%5B1%20%2B%20%5Ctan%5E%7B2%7Dx%5D%20%5Ctan%5E%7B2%7D%20x)
{Since 
}
= ![\sec^{2}x [\tan^{2}x + \tan^{4}x ]](https://tex.z-dn.net/?f=%5Csec%5E%7B2%7Dx%20%5B%5Ctan%5E%7B2%7Dx%20%2B%20%5Ctan%5E%7B4%7Dx%20%5D)
= Right hand side (Proved)
79. Left hand side
= 
= ![\cos x[1 - \sin^{2} x] \sin^{2} x](https://tex.z-dn.net/?f=%5Ccos%20x%5B1%20-%20%5Csin%5E%7B2%7D%20x%5D%20%5Csin%5E%7B2%7D%20x)
{Since 
}
= ![[\sin^{2}x - \sin^{4}x] \cos x](https://tex.z-dn.net/?f=%5B%5Csin%5E%7B2%7Dx%20-%20%5Csin%5E%7B4%7Dx%5D%20%5Ccos%20x)
= Right hand side
80. Left hand side
= 
= ![[\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x](https://tex.z-dn.net/?f=%5B%5Csin%5E%7B2%7Dx%20%2B%20%5Ccos%5E%7B2%7Dx%5D%5E%7B2%7D%20-%202%5Csin%5E%7B2%7D%20x%20%5Ccos%5E%7B2%7Dx)
{Since }
= ![1 - 2\cos^{2} x[1 - \cos^{2}x ]](https://tex.z-dn.net/?f=1%20-%202%5Ccos%5E%7B2%7D%20x%5B1%20-%20%5Ccos%5E%7B2%7Dx%20%5D)
= 
= Right hand side. (Proved)
Answer:
x= 11/5 OR 2.2
Step-by-step explanation:
