C Ruth
You can get this answer by multiplying all three numbers in each row together. After that you can see Ruth is the highest.
Answer:
i√19
Step-by-step explanation:
Because √(-19) = √(-1)√19, and because √(-1) = i, we get i√19
Answer:
4.5
Step-by-step explanation:
The square root of two is irrational. It means that you can't write it as a fraction:
![\sqrt{2} \neq \cfrac{p}{q}\quad\forall p \in \mathbb{Z},\ q \in \mathbb{Z}\setminus\{0\}](https://tex.z-dn.net/?f=%20%5Csqrt%7B2%7D%20%5Cneq%20%5Ccfrac%7Bp%7D%7Bq%7D%5Cquad%5Cforall%20p%20%5Cin%20%5Cmathbb%7BZ%7D%2C%5C%20q%20%5Cin%20%5Cmathbb%7BZ%7D%5Csetminus%5C%7B0%5C%7D%20)
Here's a proof. Assume that you could write
![\sqrt{2} = \cfrac{p}{q}](https://tex.z-dn.net/?f=%20%20%5Csqrt%7B2%7D%20%3D%20%5Ccfrac%7Bp%7D%7Bq%7D%20)
for some integers p and q, and that p and q have no common divisors. If you square both sides you have
![2 = \cfrac{p^2}{q^2} \implies p^2 = 2q^2](https://tex.z-dn.net/?f=%202%20%3D%20%5Ccfrac%7Bp%5E2%7D%7Bq%5E2%7D%20%5Cimplies%20p%5E2%20%3D%202q%5E2%20)
So,
is even, which means that also p is even, and thus there exists some number k such that
. The expression becomes
![p^2 = 2q^2 \implies 4k^2 = 2q^2 \iff q^2 = 2k^2](https://tex.z-dn.net/?f=%20p%5E2%20%3D%202q%5E2%20%5Cimplies%204k%5E2%20%3D%202q%5E2%20%5Ciff%20q%5E2%20%3D%202k%5E2%20)
So, also
is even, and thus q is also even.
But then, p and q are both even, whereas we assumed that they had no common divisors. Contraddiction.
As for the value, any calculator will give you an approximation that starts with
![\sqrt{2} = 1.4142\ldots](https://tex.z-dn.net/?f=%20%5Csqrt%7B2%7D%20%3D%201.4142%5Cldots%20)
so, it is best approximated by 1.4.
Answer:
We can't see your answer choices, but it would be n + d = 368 and .05n + .10d = 28.40
Step-by-step explanation: