All categories

3 years ago

Please help!!!!!!!!! I'll mark brainliest if you give an explanation!!!!

Mathematics

1 answer:

Maru [420]3 years ago

5 0

Answer:

say whaaaaaaaaaaaaaaàaaaaaaaaaaaaattttt??? i dont understand

You might be interested in

Use a Maclaurin series to obtain the Maclaurin series for the given function.

Rama09 [41]

Answer:

$14x cos(\frac{1}{15}x^{2})=14 \sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}$

Step-by-step explanation:

In order to find this Maclaurin series, we can start by using a known Maclaurin series and modify it according to our function. A pretty regular Maclaurin series is the cos series, where:

$cos(x)=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}$

So all we need to do is include the additional modifications to the series, for example, the angle of our current function is: $\frac{1}{15}x^{2}$ so for

$cos(\frac{1}{15}x^{2})$

the modified series will look like this:

$cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15}x^{2})^{2k}}{(2k)!}$

So we can use some algebra to simplify the series:

$cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15^{2k}}x^{4k})}{(2k)!}$

which can be rewritten like this:

$cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}$

So finally, we can multiply a 14x to the series so we get:

$14xcos(\frac{1}{15}x^{2})=14x\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}$

We can input the x into the series by using power rules so we get:

$14xcos(\frac{1}{15}x^{2})=14\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}$

And that will be our answer.

3 0

3 years ago

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

Answer:

(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

$M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}$

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

6 0

4 years ago

Which could be the name of of a point?<br> A.)1<br> B.)M<br> C.)t<br> D.)m

pishuonlain [190]

B.)M is the correct answer I thons , because mostly we use capital letters for naming points.

5 0

4 years ago

Please help and leave a small explanation!

Travka [436]

The answer is C because A and X are in the same place.

4 0

3 years ago

Read 2 more answers

Determine whether each pair of triangles is similar. If similarity exists, write a congruency statement relating to the two tria

taurus [48]

To be similar ratios of the corresponding sides should be constant.

|UV|/|EO| = 8/4=2/1

|VL|/|OG| = 4/2=2/1

|LU|/|GE| = 6/3 = 2/1

So,all three corresponding pair of these triangles UVL and EOG are in proportion, so ΔUVL and ΔEOG are similar.

Because ΔUVL and ΔEOG are similar, their corresponding angles are congruent.

m∠U=m∠E

m∠V=m∠O

m∠L=m∠G.

5 0

3 years ago

Read 2 more answers