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kogti [31]
2 years ago
6

Write 3c- 11c^4 - 6 + 5c^2 in standard form. Write degree of the polynomial.

Mathematics
1 answer:
Alborosie2 years ago
3 0

Answer:

−11c4+5c2+3c−6  degree is 4

Step-by-step explanation:

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Measurement A equals 6X - 35°
marissa [1.9K]

Answer:

............the answer is 97

6 0
2 years ago
It wont let me post a picture of the angles but they are verticle angles and the equations are (3x+8), (5x-40) and it says find
____ [38]
Vertical angles are equal

5x - 40 = 3x + 8
5x - 3x = 8 + 40
2x = 48
x = 48/2
x = 24

m∠ABC = 3x + 8 = 3 * 24 + 8 = 80°

(or m∠ABC = 5x - 40 = 5 * 24 - 40 = 80°)
8 0
2 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
The graph represents a map where each unit is 0.75 miles. Use the map to answer the question. Round to the nearest hundredth, if
prohojiy [21]

Answer:

Try 11.25 miles

Step-by-step explanation:

6 0
2 years ago
Hi pls help I suck at math
Lelechka [254]

Step-by-step explanation:

93.3

this time will help

5 0
2 years ago
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