Answer:
I think ( b) is non proportional linear
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
Answer:
1806 seats.
Step-by-step explanation:
From the question given above, the following data were obtained:
Row 1 = 24 seats
Row 2 = 27 seats
Row 3 = 30 seats
Total roll = 28
Total number of seat =?
From the above data, we can liken the roll to be in arithmetic progress.
Also, we are asked to determine the total number of seats in the theater.
Thus the sum of the sequence can be written as:
Roll 1 + Roll 2 + Roll 3 +... + Roll 28 i.e
24 + 27 + 30 +...
Thus, we can obtain obtained the total number of seats in the theater by applying the sum of arithmetic progress formula. This can be obtained as follow:
First term (a) = 24
Common difference (d) = 2nd term – 1st term
Common difference (d) = 27 – 24 = 3
Number of term (n) = 28
Sum of the 28th term (S₂₈) =?
Sₙ = n/2 [2a + (n –1)d]
S₂₈ = 28/2 [2×24 + (28 –1)3]
S₂₈ = 14 [48 + 27×3]
S₂₈ = 14 [48 + 81]
S₂₈ = 14 [129]
S₂₈ = 1806
Thus, the number of seats in the theater is 1806.
Think of it this way: your friend is thinking of a year, and you have to guess which. She has told you the year is in the 20th century (1900s).
There are 100 years in the 20th century (1901-1999 plus 1900).
There can be 100 possibilities
