9514 1404 393
Answer:
n² -n +1
Step-by-step explanation:
First differences of the terms of the sequence are ...
3-1 = 2
7-3 = 4
13-7 = 6
21-13 = 8
Second differences are ...
4-2 = 2
6-4 = 2
8-6 = 2
In a quadratic sequence the coefficient of the squared term is half of the second difference value. Here, that means the squared term will be (2/2)n² = n².
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We can find the other terms of the quadratic expression by considering the differences between n² and the actual sequence.
The sequence of n² terms is ...
1, 4, 9, 16, 25
When we subtract these from the actual sequence, we get ...
1-1 = 0
3-4 = -1
7-9 = -2
13-16 = -3
21-25 = -4
That is, the amount subtracted from n² is one less than the term number.
The expression for the n-th term is ...
an = n² -n +1
²-1
We will investigate the behavior of
both the numerator and denominator of h(x) near the point where x = 1. Let
f(x)= x³ + x -2 and g(x)=x²-1. Find the local linearizations of f and g at a = 1,
and call these functions Lf(x) and Lg(x), respectively.
Lf(x) =
L₂(x) =
Explain why h(x) ≈
Lf(x)
Lg(x)
for a near a = 1.
The local linearizations of f(x) and g(x) at a = 1 are Lf(x) = 4x - 5 and Lg(x) = 2x - 2 respectively. The function h(x) ≈ Lf(x)/Lg(x) because the local linearizations provide a good approximation of the numerator and denominator of h(x) near x = 1.
The local linearization of a function at a given point is an approximation of the function using a linear equation. To find the local linearization of a function f at a = 1, we need to find the slope of the tangent line at a = 1, which is equivalent to finding the derivative of f at x = 1. By taking the derivative of f(x) = x³ + x - 2, we get f'(x) = 3x² + 1. Evaluating f'(1), we find that the slope of the tangent line at a = 1 is 4. Therefore, the local linearization of f at a = 1, denoted as Lf(x), is given by Lf(x) = f(a) + f'(a)(x - a), which becomes Lf(x) = -1 + 4(x - 1) = 4x - 5.
Similarly, to find the local linearization of g(x) = x² - 1 at a = 1, we need to find the slope of the tangent line at a = 1. The derivative of g(x) is g'(x) = 2x. Evaluating g'(1), we find that the slope of the tangent line at a = 1 is 2. Therefore, the local linearization of g at a = 1, denoted as Lg(x), is given by Lg(x) = g(a) + g'(a)(x - a), which becomes Lg(x) = 0 + 2(x - 1) = 2x - 2.
When investigating the behavior of the function h(x) = (f(x))/(g(x)) near the point x = 1, we can approximate h(x) using the local linearizations of f and g at a = 1. Near the point a = 1, h(x) ≈ Lf(x)/Lg(x) because Lf(x) and Lg(x) provide a good approximation of the numerator and denominator, respectively, of h(x). This approximation holds as long as x is close to 1.
130 feet
PLEASE ANSWER I REALLY NEED IT
Answer:
it is at 210 feet
Step-by-step explanation:
it rises 340 feet, then it descends 130 so we subtract 130 from 340 which gets us 210 feet.
k=9 soo got 9 by adding 6 and 8 which equals 14 then i thought 14 minus what equals 5 and got 9 which is the answer for k.
Answer:
c There is sufficient evidence to conclude that the proportion of high school students stayed 0.096 at this counselor's high school O
Step-by-step explanation:
Given that according to the Centers or Disease Control and Prevention, 9.6% of high school students current through (c) below a
(a) Determine the null and alternative hypotheses.
(right tailed test for proportion of high school students )
b) If the null hypothesis should not be rejected, state the conclusion of the high school counselor.
c There is sufficient evidence to conclude that the proportion of high school students stayed 0.096 at this counselor's high school O