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3 years ago

Someone can help??

Mathematics

1 answer:

oee [108]3 years ago

4 0

Answer:

Step-by-step explanation:

let cos^{-1}x=t

cos t=x

when x=1,cos t=1=cos 0

t \rightarrow 0

$\lim_{x \to 1} \frac{1-\sqrt{x}}{(cos ^{-1}x)^2 } \\= \lim_{t \to 0} \frac{1-\sqrt{cos~t}}{t^2} \times \frac{1+\sqrt{cos~t}}{1+\sqrt{cos ~t}} \\= \lim_{t \to 0} \frac{1-cos~t}{t^2(1+\sqrt{cos~t})}} \\= \lim_{t \to 0 }\frac{2 sin^2~\frac{t}{2}}{t^2(1+\sqrt{cos~t})}} \\= 2\lim_{t \to 0 }(\frac{sin~t/2}{\frac{t}{2} })^2 \times \frac{1}{4} \times \lim_{t \to 0 }\frac{1}{1+\sqrt{cos~t}} \\=\frac{2}4} \times 1^2 \times \frac{1}{1+\sqrt{cos~0}} \\=\frac{1}{2} \times \frac{1}{1+1} \\=\frac{1}{4}$

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Simplify.

marin [14]

1/2+(3/8−1/8)^2

= 1/2+(2/8)^2

= 1/2+(1/4)^2

= 1/2 + 1/16

= 8/16 + 1/16

= 9/16

Answer

9/16

3 0

2 years ago

Read 2 more answers

The sum of the interior angles of each polygon is 360°. In the diagram, DEFG=SPQR.Find the values of x and y

velikii [3]

Never gonna give you up
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4 0

2 years ago

The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses.

Luden [163]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

If we had increased the confidence level, the margin of error also would have increased.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96. Increasing the confidence level, z also increases, hence the margin of error also would have increased.

The sample size and the estimate are given as follows:

$n = 450, \pi = 0.47$ .

The lower and the upper bound of the interval are given, respectively, by:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.4239$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.5161$

The 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).

More can be learned about the z-distribution at brainly.com/question/25890103

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5 0

1 year 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

There is a bag with only red marbles and blue marbles.

hram777 [196]

Answer:

Step-by-step explanation:

The probability of choosing a blue marble is 7/8, or 49/56, which means that out of 56 marbles, 49 are blue. 56-49=7, so there are 7 red marbles. Hope this helps! :)

7 0

2 years ago

Someone can help?? ​

Someone can help??