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3 years ago

5

Determine the lateral and total surface

1 answer:

likoan [24]3 years ago

8 0

The answer to your question would be 16

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If y = 5x − 4, which of the following sets represents possible inputs and outputs of the function represented as ordered pairs?

zzz [600]

The answer is not in the choices so ... but the (0,4), (1,1), (2,6)

Hope This Help.

Xd <3

6 0

3 years ago

What is the equation for 5 units left and 3 units up from f(x)=x

erastovalidia [21]

The new equation after shifting will be:

$g(x) = (x+5)+3\\g(x) = x + 8$

Step-by-step explanation:

Function trnafomations upward and left are defined as:

Upward:

f(x) => f(x)+b where b is an integer

Left:

f(x) => f(x+b) where b is an integers

Given function is:

$f(x) = x$

Shifting the function 5 units left

$g(x) = f(x+5) => x+5\\g(x) = x+5$

Shifting the function upward 2 units

So,

$g(x) = (x+5)+3\\g(x) = x + 8$

The new equation after shifting will be:

$g(x) = (x+5)+3\\g(x) = x + 8$

Keywords: Functions, shifting

Learn more about functions at:

#LearnwithBrainly

3 0

3 years ago

Which number is greater than square root of one hundred fifty five?

joja [24]

Answer:

try 3

Step-by-step explanation:

5 0

3 years ago

A random sample of 40 students from each grade level was surveyed regarding their preference for a class trip. If there are 220

NNADVOKAT [17]

Answer:

66 students prefer the zoo

Step-by-step explanation:

I got it correct :)

3 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