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Maru [420]
2 years ago
8

Write the equation of a line that passes through the points in the table.

Mathematics
1 answer:
lord [1]2 years ago
6 0

Answer: do it have a answer choice if it do then its B,C and e

Step-by-step explanation:

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Please help me with 1-4 and 6 and 7
Tom [10]
I could only get number one which was; 1.75
3 0
3 years ago
Jose Rivera purchased 7 textbooks for his classes at River Run Community College. The total cost of the texts was $208.55. What
shepuryov [24]
In the question, there are certain information's that are of immense importance in regards to finding the answer. The first information is that Jose Rivera had purchased 7 textbooks for his classes at River Run Community College. The price that Jose Rivera had to pay was $208.55. The average cost per textbook needs to be found.
The cost of 7 textbooks purchased by Jose Rivera = 208.55 dollars
Then
The cost of 1 textbook purchased by Jose Rivera = (208.55/7) dollars
                                                                               = 29.79 dollars
The average cost of a textbook purchased by Jose Rivera rounded to the nearest cent is 29.80 dollars.
6 0
3 years ago
What is f(–3) for f(x) = 2x2 – x + 5?<br> A) 11 <br> B) 17 <br> C) 20 <br> D) 26
nirvana33 [79]

Answer: D

<u>Step-by-step explanation:</u>

f(x) = 2x² - x + 5

f(-3) = 2(-3)² - (-3) + 5

      = 2(9) + 3 + 5

      = 18 + 3 + 5

      = 26

6 0
3 years ago
Two identical number cubes are shown in the picture below. The edge length of these number cubes is 4 centimeters: What is the c
meriva

Answer:

d. 128 cm^3

Step-by-step explanation:

to find the volume its length time width times heights; so 4 cm times 4 cm time 4 cm which is 64. Since the dice are the same size you add 64 cm^3 to 64 cm^3 to get 128 cm^3

8 0
3 years ago
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
FromTheMoon [43]

Answer:

The Taylor series is \ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

The radius of convergence is R=3.

Step-by-step explanation:

<em>The Taylor expansion.</em>

Recall that as we want the Taylor series centered at a=3 its expression is given in powers of (x-3). With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of \ln(1+x).

Then,

\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).

Now, in order to make a more compact notation write \frac{x-3}{3}=y. Thus, the above expression becomes

\ln(x) = \ln 3 + \ln(1+y).

Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,  

\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.

Now, substitute \frac{x-3}{3}=y in the previous equality. Thus,

\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

<em>Radius of convergence.</em>

We find the radius of convergence with the Cauchy-Hadamard formula:

R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},

Where a_n stands for the coefficients of the Taylor series and R for the radius of convergence.

In this case the coefficients of the Taylor series are

a_n = \frac{(-1)^{n+1}}{ n3^n}

and in consequence |a_n| = \frac{1}{3^nn}. Then,

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}

Applying the properties of roots

\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.

Hence,

R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}

Recall that

\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.

So, as R^{-1}=\frac{1}{3} we get that R=3.

8 0
3 years ago
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