The require set (CUD)' and C'nD is {q, r, x, y} and Ø.
<h3>Set theory</h3>
Sets are elements arranged in a specific pattern or way. Given the expression below;
U= {f, k, q, r, s, x, y }
C={f, s,y}
D = {f,x)
CUD is the union of two sets and given as:
C U D = {f, s, y, k}
(CUD)' = {q, r, x, y}
For the set C'nD
C' = {k, q, r,x}
C' n D = Ø.
Hence the require set (CUD)' and C'nD is {q, r, x, y} and Ø.
Learn more on set theory here: brainly.com/question/13458417
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9514 1404 393
Answer:
a) 4
b) 3
Step-by-step explanation:
a. The total number of real and complex zeros is equal to the degree of the polynomial. That total is (1 negative real) + (3 positive real/complex) = 4 total zeros. The degree of the polynomial is 4.
The even degree is confirmed by the answer to part b, and by the end-behavior shown in the table, which has a tendency to -∞ for |x|→∞.
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b. The intermediate value theorem tells you there will be zeros in the intervals (0, 1), (1, 2), and (2, 3) according to the values in the table. (The function changes sign in those intervals.) Thus there are 3 positive real zeros.
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<em>Additional comment</em>
Stanley cannot tell anything about Descartes' rule of signs by analyzing the table of function values. To use that rule, he must have terms of the polynomial. If he has those terms, he already knows the degree of the polynomial.
THE ANSWER IS 2000 i think
Answer:
7/9
Step-by-step explanation:
this equation is true:
21/27 = 7/9
If the numerator is greater than or equal to the denominator of a fraction, then it is called an improper fraction. In that case, you could convert it into a whole number or mixed number fraction.
7/9 = Proper Fraction
