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ikadub [295]
3 years ago
8

Help!!! I don’t understand how to do this

Mathematics
1 answer:
lidiya [134]3 years ago
4 0

Answer:

  f^{(k)}(x)=\dfrac{17k!(-1)^k}{(x-9)^{k+1}}

Step-by-step explanation:

The question presumes you have access to a computer algebra system. The one I have access to provided the output in the attachment. The list at the bottom is the list of the first four derivatives of f(x).

__

The derivatives alternate signs, so (-1)^k will be a factor.

The numerators start at 17 and increase by increasing factors: 2, 3, 4, indicating k! will be a factor.

The denominators have a degree that is k+1.

Putting these observations together, we can write an expression for the k-th derivative of f(x):

  \boxed{f^{(k)}(x)=\dfrac{17k!(-1)^k}{(x-9)^{k+1}}}

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Vanessa bought a house for $268,500. She has a 30 year mortgage with a fixed rate of 6.25%. Vanessa’s monthly payments are $1,59
Genrish500 [490]

Answer:

Option a - $9,314.45

Step-by-step explanation:

Cost of the house = $268,500

Time of repayment = 30 years

Repayment is done monthly, so number of repayments = 30 X 12 = 360

Monthly Payment = $1595.85

Rate of interest per payment period = \frac{.0625}{12}

So, Present value of monthly payments = 1595.85 X  \frac{(1+\frac{.0625}{12})^{360}-1}{(1+\frac{.0625}{12})^{360}*(\frac{.0625}{12})}

= $259,185.55

So, Vanessa's down payment = $268,500 - $259,185.55 = $9,314.45

Hope it helps.

Thank you !!

8 0
3 years ago
Which of the following is an improper integral?
guapka [62]

Answer:

A)  \displaystyle \int\limits^3_0 {\frac{x + 1}{3x - 2}} \, dx

General Formulas and Concepts:

<u>Calculus</u>

Discontinuities

  • Removable (Hole)
  • Jump
  • Infinite (Asymptote)

Integration

  • Integrals
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  • Integration Constant C
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Step-by-step explanation:

Let's define our answer choices:

A)  \displaystyle \int\limits^3_0 {\frac{x + 1}{3x - 2}} \, dx

B)  \displaystyle \int\limits^3_1 {\frac{x + 1}{3x - 2}} \, dx

C)  \displaystyle \int\limits^0_{-1} {\frac{x + 1}{3x - 2}} \, dx

D) None of these

We can see that we would have a infinite discontinuity if x = 2/3, as it would make the denominator 0 and we cannot divide by 0. Therefore, any interval that includes the value 2/3 would have to be rewritten and evaluated as an improper integral.

Of all the answer choices, we can see that A's bounds of integration (interval) includes x = 2/3.

∴ our answer is A.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit:  Integration

Book: College Calculus 10e

6 0
3 years ago
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343t^6

Words to take up 20 characters
7 0
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stiv31 [10]

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And obviously, you have 4^2=16>4

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But since not all rational numbers are between 0 and 1, the general claim is false.

6 0
3 years ago
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Answer:

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