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3 years ago

If ABC is 160°, what is DBC?

Mathematics

1 answer:

Monica [59]3 years ago

5 0

The answer should be A 160

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A car travels at an average speed of 56 miles per hour. How many miles does it travel in 4 hours and 15 minutes?

ivolga24 [154]

56*(4+15/60)

56*(240+15)/60

56*(255)/60

238 miles

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3 years ago

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Pls show working 3 friends shared the driving on a long trip. Marla drove 7 miles more than Guido. Guido drove five times as far

olga2289 [7]

Step-by-step explanation:

Marla drove 567

Gudio drove 560

The trip was 679

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2 years ago

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What angle of rotation does the letter s have? a. 0° b. 90° c. 120° d. 180°

anyanavicka [17]

I think it is d.180°

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3 years ago

Write the solution as an ordered pair (x,y)

garik1379 [7]

Answer:

(-2, -3)

Step-by-step explanation:

2x + 1 = 3x + 3

minus 3 from both sides

2x - 2 = 3x

minus 2x from both sides

-2 = x

then plug in -2 for x

y = 2(-2) + 1

y = -4 + 1

y = -3

(-2, -3).

8 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