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son4ous [18]
4 years ago
14

What are the solutions of the system solve by graphing

Mathematics
1 answer:
Vladimir [108]4 years ago
7 0
Vertex of x^2-2x-1 is (1,-2) so B is true
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Cali bought a sweater priced a $60. If the sales tax is 9% what is the total purchase price of the sweater?
forsale [732]

Answer:

65.4

Step-by-step explanation:

60*.09 = 5.4

60+5.4 = 65.4

8 0
3 years ago
135/11 improper fraction to a mixed number
Scorpion4ik [409]

Answer:

12 \frac{3}{11}

5 0
3 years ago
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Hugh bought some magazines that cost $3.95 each and some books that cost $8.95 each. He spent a total of $47.65. If Hugh bought
DiKsa [7]
So first you want to find the price of all the magazines he bought which is 3.95 x 3 = 11.85.
then you are going to find out how much money he had left to spend on books which is 47.65-11.85=35.80 

then you want to find how many books he bought with that money so you would take 35.80/8.95= 4

so he bought 4 books
8 0
4 years ago
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SLOVE <br> 18(5) <br><br> A. 95 <br><br> B. 90<br><br> C. -90<br><br> D. -95
Evgesh-ka [11]

Answer:

90 i did the math

plzz mark brainllest

6 0
3 years ago
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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
4 years ago
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