Let U = {1, 2, 3, 4, 5, 6, 7}, A= {1, 3, 4, 6}, and B= {3, 5, 6}. Find the set A’ U B’
Art [367]
Answer:
Step-by-step explanation:
A'={2,5,7}
B'={1,2,4,7}
A'UB'={1,2,4,5,7}
Answer: ok bet
Step-by-step explanation: theres a picture below and to graph it you simply start with the y coordinate, which is 7. you put 7 on the graph then go down 3 because 3 is negative, then go right one because 1 is postitive. (you get 1 from making the slope into a fraction, -3/1) also remember the formula for this is y=mx+b. its tricky but i hope this helped
Answer:
Step-by-step explanation:
The first differences of the sequence are ...
- 5-2 = 3
- 10-5 = 5
- 17-10 = 7
- 26-17 = 9
- 37-26 = 11
Second differences are ...
- 5 -3 = 2
- 7 -5 = 2
- 9 -7 = 2
- 11 -9 = 2
The second differences are constant, so the sequence can be described by a second-degree polynomial.
We can write and solve three equations for the coefficients of the polynomial. Let's define the polynomial for the sequence as ...
f(n) = an^2 + bn + c
Then the first three terms of the sequence are ...
- f(1) = 2 = a·1^2 + b·1 + c
- f(2) = 5 = a·2^2 +b·2 + c
- f(3) = 10 = a·3^2 +b·3 +c
Subtracting the first equation from the other two gives ...
3a +b = 3
8a +2b = 8
Subtracting the first of these from half the second gives ...
(4a +b) -(3a +b) = (4) -(3)
a = 1 . . . . . simplify
Substituting into the first of the 2-term equations, we get ...
3·1 +b = 3
b = 0
And substituting the values for a and b into the equation for f(1), we have ...
1·1 + 0 + c = 2
c = 1
So, the formula for the sequence is ...
f(n) = n^2 + 1
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The 20th term is f(20):
f(20) = 20^2 +1 = 401
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<em>Comment on the solution</em>
It looks like this matches the solution of the "worked example" on your problem page.
40 of the toys weren't dolls. The reason why is 5/8 times 64 is equal to 40.
Let's say the numbers are "a" and "b"
thus
set the derivative to 0, and check the critical points, there's only one anyway
and do a first-derivative test, to see if it's a maximum
