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evablogger [386]
4 years ago
15

Consider the error in using the approximation e−x≈1−x on the interval [−1,1]. (a) reasoning informally, on what interval is this

approximation an overestimate? 1 an underestimate? (for each, give your answer as an interval or list of intervals,
e.g., to specify the intervals −0.25≤x<0.5 and 0.75
Mathematics
1 answer:
crimeas [40]4 years ago
6 0

e^{-x} \approx 1 - x

is a classic approximation, true for small x.

The next term in the polynomial expansion will be +kx^2 where k is a positive number. So our estimate 1-x is definitely an underestimate on both sides of x=0.

Since for negative xs the exponential rises exponentially and the line only linearly, the exponential exceeds the line for all negative x. For positive x, the line quickly goes negative while the exponential is always positive.

So, there's no interval for which our approximation is an overestimate.

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Find the next two terms in the pattern A,A,B,A,C,A,D,A
andrey2020 [161]

Answer:

E, A, F, A

Step-by-step explanation:

A,A

B,A

C,A

D,A

E,A

F,A

6 0
3 years ago
What’s the remainder for this synthetic division problem?
Vesna [10]

3   |   1    5     -8     6

.    |         3    24   48

- - - - - - - - - - - - - - - -

.    |    1    8    16    54

That is to say,

\dfrac{x^3+5x^2-8x+6}{x-3}=x^2+8x+16+\dfrac{54}{x-3}

The remainder is 54.

Another way of doing it is to apply the polynomial remainder theorem, which says the remainder upon dividing a polynomial p(x) by x-c is exactly p(c). So recognizing that the listed coefficients refer to

p(x)=x^3+5x^2-8x+6

we find

p(3)=3^3+5\cdot3^2-8\cdot3+6=54

7 0
4 years ago
The interest rate r required to increase your investment p to the amount a in t years is found by . Find the interest rate r for
Dahasolnce [82]

Answer:

The interest rate was of 0.1173 = 11.73%.

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

E = P*I*t

In which E is the amount of interest earned, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

After t years, the total amount of money is:

A = E + P

In this problem, we have that:

A = 10000, P = 8100, t = 2

So

A = E + P

10000 = E + 8100

E = 1900

So

1900 = 8100*I*2

I = \frac{1900}{8100*2}

I = 0.1173

The interest rate was of 0.1173 = 11.73%.

7 0
3 years ago
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