Answer:
Hi, there your answer is C or D and the second page is C
Step-by-step explanation:
Since it's not a domain restricted function anywhere, the domain of this function is (-∞, ∞). Because the x is squared, you will <em>never </em>get a negative answer, and therefore, the range is either (0, ∞) or {x: x > 0}. Hope this helps you out!
Answer:
See explanation below
Step-by-step explanation:
To prove by contradiction, we are going to assume that a and b are odd.
If a and b are odd, then there exists integers j, k > 0 such that
a = 2j + 1 and b= 2k + 1
We're going to rewrite the original expression substituting a and b by their odd expression.
![a^{2} +b^{2} = (2j+1)^{2} +(2k+1)^{2} =4j^{2} +4j+1+4k^{2} +4k+1=4j^{2} +4k^{2} +4j+4k+2\\=4(j^{2} +k^{2}+j+k) +2](https://tex.z-dn.net/?f=a%5E%7B2%7D%20%2Bb%5E%7B2%7D%20%20%3D%20%282j%2B1%29%5E%7B2%7D%20%2B%282k%2B1%29%5E%7B2%7D%20%3D4j%5E%7B2%7D%20%2B4j%2B1%2B4k%5E%7B2%7D%20%2B4k%2B1%3D4j%5E%7B2%7D%20%2B4k%5E%7B2%7D%20%2B4j%2B4k%2B2%5C%5C%3D%3Cstrong%3E4%28j%5E%7B2%7D%20%2Bk%5E%7B2%7D%2Bj%2Bk%29%20%2B2%3C%2Fstrong%3E)
Now we have to cases, c is even or c is odd.
Case 1: If c is odd.
If c is odd, then c² is also odd, but we have that the expression above is even. Therefore, this is a contradiction.
Case 2: If c is even.
If c is even then it's multiple of 2, and c² is multiple of 4, but the expression above is not multiple of 4 (because it has the form 4g + 2). Therefore we have a contradiction.
Thus, a or b must be even.
Answer:
5.19
Step-by-step explanation:
first do 28.86 / 6 to find out how much each paid. then subtract that number (4.81) from 10
Answer:
a. 5/17
b.12/17
c.0/17
d.17/17
Step-by-step explanation: