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:
No extraneous solution
Step-by-step explanation:
We have the logarithmic equation given by,
![\log_{2}[\log_{2}(\sqrt{4x})]=1](https://tex.z-dn.net/?f=%5Clog_%7B2%7D%5B%5Clog_%7B2%7D%28%5Csqrt%7B4x%7D%29%5D%3D1)
i.e. 
i.e. 
i.e. 
i.e. 
i.e. 
i.e. 
So, the solution of the given equation is x=4.
Now, as we domain of square root function is x > 0 and also, the domain of logarithmic function is
.
Therefore, the domain of the given function is x > 0.
We know that the extraneous solution is the solution which does not belong to the domain.
But as x=4 belongs to the domain x > 0.
Thus, x = 4 is not an extraneous solution.
Hence, this equation does not have any extraneous solution.
Yes the absolute value of 2 is 2!
Answer:
b. n-4
Step-by-step explanation:
a number=n
4 less than it =n-4
With an inequality you have to be careful because you never know when you are going to do the wrong steps in the equation