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3 years ago

I need help I can’t figure this out

Mathematics

1 answer:

adelina 88 [10]3 years ago

5 0

Answer:

12.04 = BC

Step-by-step explanation:

A tangent intersects at 90 degrees

We know AB and AC

We can use the Pythagorean theorem

a^2 + b^2 = c^2

AC^2 + AB^2 = BC^2

8^2 + 9^2 = BC^2

64+ 81 = BC^2

145 = BC^2

Take the square root of each side

sqrt(145) = BC^2

12.0415945 BC

To the nearest hundredth

12.04 = BC

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Write y=1/6x+7 in standard form using integers

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The standard form using integers of the equation is 6y - x = 42

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

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I suppose you mean

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which converges everywhere. Then by substitution,

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b. Naturally, the differentiated series represents

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

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

Multiplying by $-2x$ gives

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

and from here,

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

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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.

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