Answer:
The theorem here is essentially that
if a and 3 are disjoint sets with
exactly one element each, then their
union has exactly two elements. ...
Peano shows that it's not hard to
produce a useful set of axioms that
can prove 1+1=2 much more easily
than Whitehead and Russell do.
Answer:
The answer is -23
Step-by-step explanation:
14 books in 3 months is 35%
all of books is 100%
14- 35%
x- 100%
we calculate from proportions, crosswise:
![x=\frac{14*100}{35}=\frac{1400}{35}=40](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B14%2A100%7D%7B35%7D%3D%5Cfrac%7B1400%7D%7B35%7D%3D40)
Answer: Benito sign up to read 40 books.
Greetings,
n00nst00p :)
Answer: 1/6
Step-by-step explanation:
The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.
Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.
So, if we sum the first N odd numbers, we have
![\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csum_%7Bi%3D1%7D%5EN%202i-1%20%3D%202%5Csum_%7Bi%3D1%7D%5EN%20i%20-%20%5Csum_%7Bi%3D1%7D%5EN%201)
The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have
![\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%5Csum_%7Bi%3D1%7D%5EN%20i%20%3D%202%5Ccdot%5Cdfrac%7BN%28N%2B1%29%7D%7B2%7D%3DN%28N%2B1%29)
The second sum is simply the sum of N ones:
![\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N](https://tex.z-dn.net/?f=%5Cunderbrace%7B1%2B1%2B1%5Cldots%2B1%7D_%7BN%5Ctext%7B%20times%7D%7D%3DN)
So, the final result is
![\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csum_%7Bi%3D1%7D%5EN%202i-1%20%3D%202%5Csum_%7Bi%3D1%7D%5EN%20i%20-%20%5Csum_%7Bi%3D1%7D%5EN%201%20%3D%20N%28N%2B1%29-N%20%3D%20N%5E2%2BN-N%20%3D%20N%5E2)
which ends the proof.