Yes. The dependent variable depends on the independent variable.
Prove that if m + n and n + p are even integers, where m, n, and p are integers, then m + p is even.
m=2k-n, p=2l-n
Let m+n and n+p be even integers, thus m+n=2k and n+p=2l by definition of even
m+p= 2k-n + 2l-n substitution
= 2k+2l-2n
=2 (k+l-n)
=2x, where x=k+l-n ∈Z (integers)
Hence, m+p is even by direct proof.
Answer:
Step-by-step explanation:
I will ASSUME you mean
-41 = (-2/5)(45n + 60) + n
and not -41 = -2/(5(45n + 60)) + n
or -41 = -2/(5(45n + 60) + n)
-41 = (-2/5)(45n + 60) + n
distribute the (-2/5)
-41 = -18n - 24 + n
combine like terms
17n = 17
reduce to simplest form
n = 1
Answer:
Step-by-step explanation:
sin2x = 2sinx*cosx
= 
* cosx
= 
* cosx
= 2 
= 
= 24/25
X + (x + 10) = 154
2 x + 10 = 154
2 x = 154-10 = 144
x = 72 (ans)
Not sure if it’s right
