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Andrew [12]
3 years ago
8

What is the distance, to the nearest tenth

Mathematics
1 answer:
Alecsey [184]3 years ago
7 0

I think the answer may be B or c but I would say c

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Ashton is saving money for a new bike. He needs 120$ But only has saved 60% so far. How much more money does he need to buy the
kherson [118]

Answer:

$48

Step-by-step explanation:

Ashton is saving money for a new bike. He needs 120$ But only has saved 60% so far. How much more money does he need to buy the scooter

Ashton needs money for new bike=120$

savings of ashton is 60 % of his bike cost

hence 60% of 120 dollar will be

His saving=(60/100)*120=7200/100=72$

remaining cost will be obtained by subtracting saved money from total cost

total cost=120 dollar

saved money=72dollar

hence

Remaining more money he needed is =120-72=48$


7 0
3 years ago
Order the numbers from least to greatest. <br> 0.5, <br> -2, <br> 1/4, <br> -6.3
adoni [48]

Answer:

Given

Step-by-step explanation:

We have:

0.5

-2.0

-6.3

1/4 = 0.25

So ordering from least to greatest the greater negative integer comes since it is to the far left on the number line to the greater positive integer. Therefore the answer is:-

-6.3, -2, 0.25, 0.5

= -6.3, -2, 1/4, 0.4

5 0
3 years ago
Read 2 more answers
11. Evaluate f (x)=- x + 2 for f( – 5) and f(1).
kenny6666 [7]

Answer:

-3 and positive 3

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
PLEASE HELP ILL GIVE BRAINLIEST
Elza [17]
The answer should be b
6 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
4 years ago
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