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2 years ago

Need help help for 30 points

Mathematics

2 answers:

Slav-nsk [51]2 years ago

8 0

100x24 because u have to count the shaded squares

S_A_V [24]2 years ago

4 0

100 times 17 bc u have to count the shdded parts

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Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

7 0

3 years ago

WILL GIVE BRAINLIEST TO PERSON WHO ANSWERS CORRECTLY!!!!!!

maria [59]

I'm pretty sure its the second one

3 0

2 years ago

May I please have help with 2-14 even on this ? I'm horrible at math.

Vilka [71]

2. x = 1/4 ( 0.25)
4. y = 122, x = 42
6. x = 63, y = 8
8. is missing
10. 13.5 cm
12. x = 10.6
14. x = 5.45

6 0

3 years ago

What is the basic difference between liability insurance and collision insurance?

tia_tia [17]

Liability insurance only covers incidents that youre liable for... so if you hit a car, your insurance pays for damages on other persons car and not your own car. Collision is usually covered in a collision with another vehicle.. so like not if you hit a tree or something

4 0

3 years ago

A 5.8-ft-tall person walks away from a 9-ft lamppost at a constant rate of 3.4 ft/sec. What is the rate that the tip of the pers

Dovator [93]

Answer:

9.56 ft/sec

Step-by-step explanation:

We are told that a 5.8-ft-tall person walks away from a 9-ft lamppost at a constant rate of 3.4 ft/sec.

I've attached an image showing triangle that depicts this;

Thus; dx/dt = 3.4 ft/sec

From the attached image and using principle of similar triangles, we can say that;

9/y = 5.8/(y - x)

9(y - x) = 5.8y

9y - 9x = 5.8y

9y - 5.8y = 9x

3.2y = 9x

y = 9x/3.2

dy/dx = 9/3.2

Now, to find how fast the tip of the shadow is moving away from the lamp post, it is;

dy/dt = dy/dx × dx/dt

dy/dt = (9/3.2) × 3.4

dy/dt = 9.5625 ft/s ≈ 9.56 ft/sec

5 0

2 years ago