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3 years ago

Using the following table, what is the correct equation?

Mathematics

1 answer:

kompoz [17]3 years ago

6 0

Answer: 50

Step-by-step explanation: hope this helpss

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A student theater charges $8.50 per ticket.

Burka [1]

Answer: A) 680

B) if the theater charges more per ticket, then they can sell fewer tickets to reach their goal of $750.

Step-by-step explanation:

A) 8.50x +70 = 750

x=80

8.50(80)+70=750

7 0

3 years ago

What are the solutions of t+3>-1? Graph the solutions.

olya-2409 [2.1K]

T is more than -4. Remember that -4 is not included.

4 0

3 years ago

What about this one ??

V125BC [204]

The first expression is equivalent to B

The second expression is equivalent to A

The third expression is equivalent to C

You are basically adding like terms. For example for the first expression, it is 7x^2 + 2x^2, which equals 9x^2. Remember since you're adding, do not add the exponents.

4 0

3 years ago

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

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

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

$-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)$

c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

The limit is 0, so any choice of $x$ satisfies the convergence condition.

3 0

3 years ago

Which graph best represent the inequality? pictures down below , Thank you all <3

kow [346]

Answer:

y ≤ 3x+1

Step-by-step explanation:

when the line is solid with no spaces in between, that automatically rules out A) and C). and the shading is underneath the line so that rules out B) as well. Also, if you look through the line that passes through the Y-axis, you can see that it is on +1 so that automatically tells you that both -1 are out.

Hope this helped you!

#MRS

6 0

1 year ago