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2 years ago

At a quick-lunch counter, 3 pretzels and 1 soda cost

Mathematics

1 answer:

NeX [460]2 years ago

8 0

Answer:

I really did try man ;-;

Step-by-step explanation:

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Can someone answer this

11111nata11111 [884]

Answer:

Well first see what the points are

Step-by-step explanation:

5 0

3 years ago

Find the volume Help ASAP pleat

andrew-mc [135]

Answer: 330 yd³

Step-by-step explanation:

v = whl

v = 6 x 11 x 5

v = 66 x 5

v = 330 yd³

7 0

2 years ago

Read 2 more answers

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

2 years ago

If an arithmetic progression satisfies U4=7 , U8=15 .Find U10

Alja [10]

Answer:

U10 = 19

Step-by-step explanation:

The sequence is $Ux=(x*2)-1$

x is the coefficient to U.

So, $U10=(10*2)-1$

Which means $U10=20-1$

Which means $U10=19$

And there is your answer.

Hope this helps. Have a nice day.

6 0

2 years ago

Plsssss helppppppp because I am v confused

Allisa [31]

The simplified version of this expression is 110g^4h^16v^6. first, remove the parenthesis. the expression becomes 10g^3h^8v^6 times 11gh^8. then, calculate the product.

4 0

2 years ago