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scZoUnD [109]
3 years ago
5

A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between

times of maximum brightness is 6.2 days. The average brightness of this star is 3.0 and its brightness changes by ±0.25. In view of these data, the brightness of the star at time t, where t is measured in days, has been modeled by the function B(t)=4.2 +0.45sin(2pit/4.4)
(a) Find the rate of change of the brightness after t days.
(b) Find the rate of increase after one day.
Mathematics
1 answer:
sattari [20]3 years ago
8 0

Answer:

a)

B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)

b) 0.09

Step-by-step explanation:

We are given the following in the question:

B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(t) = 0.45\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\times \dfrac{2\pi}{4.4}\\\\B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)

b) rate of increase after one day.

We put t = 1

B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09

The rate of increase after 1 day is 0.09

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Find the two geometric means between 20 and 2500
iris [78.8K]

Answer:

<h2>The two geometric means between 20 and 2500 are <em>100 and 500.</em></h2>

Step-by-step explanation:

First of all we all should know about a <em>geometric progression </em>to solve this question.

A geometric progression is a series in which there is a first term <em>a </em>and all the next terms are calculated by multiplying the previous term by a common number <em>r</em>.

where <em>a</em> is known as first term and

<em>r</em> is known as common ratio.

In the question we are given <em>a </em>as 20 and we have to find out 2 terms after 20 and 4th term is given as 2500.

Formula for n^{th} term in a geometric progression is:

a_{n}  = a\times r^{n-1}

Here a_{4} = 2500

As per formula of n^{th} term:

a\times r^{3} = 2500\\\Rightarrow 20 \times r^{3} =2500\\\Rightarrow r^{3} = 125\\\Rightarrow r = 5

Now, 2nd term:

a_{2} = a \times r\\\Rightarrow a_{2} = 20 \times 5\\\Rightarrow a_{2} = 100

Now, 3rd term:

a_{3} = a \times r^{2} \\\Rightarrow a_{3} = 20 \times 5^{2}\\\Rightarrow a_{3} = 500

So, the two geometric means between 20 and 2500 are <em>100 and 500</em>.

6 0
3 years ago
Integral of x"2+4/x"2+4x+3
dolphi86 [110]

I'm guessing you mean

\displaystyle \int\frac{x^2+4}{x^2+4x+3}\,\mathrm dx

First, compute the quotient:

\displaystyle \frac{x^2+4}{x^2+4x+3} = 1 + \frac{4x-1}{x^2+4x+3}

Split up the remainder term into partial fractions. Notice that

<em>x</em> ² + 4<em>x</em> + 3 = (<em>x</em> + 3) (<em>x</em> + 1)

Then

\displaystyle \frac{4x-1}{x^2+4x+3} = \frac a{x+3} + \frac b{x+1} \\\\ \implies 4x - 1 = a(x+1) + b(x+3) = (a+b)x + a+3b \\\\ \implies a+b=4 \text{ and }a+3b = -1 \\\\ \implies a=\frac{13}2\text{ and }b=-\frac52

So the integral becomes

\displaystyle \int \left(1 + \frac{13}{2(x+3)} - \frac{5}{2(x+1)}\right) \,\mathrm dx = \boxed{x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C}

We can simplify the result somewhat:

\displaystyle x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C \\\\ = x + \frac12 \left(13\ln|x+3| - 5\ln|x+1|\right) + C \\\\ = x + \frac12 \left(\ln\left|(x+3)^{13}\right| - \ln\left|(x+1)^5\right|\right) + C \\\\ = x + \frac12 \ln\left|\frac{(x+3)^{13}}{(x+1)^5}\right| + C \\\\ = \boxed{x + \ln\sqrt{\left|\frac{(x+3)^{13}}{(x+1)^5}\right|} + C}

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S greater than or equal to 10
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Hope this helped :)

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