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tankabanditka [31]
3 years ago
14

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


8 0
3 years ago
Read 2 more answers
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

5 0
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°
8 0
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
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