MATHEMATICS COLLEGE

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 4.2 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) = 3.0 + 0.25 sin 2πt 4.2 . (a) Find the rate of change of the brightness after t days. dB dt =

Answers

Answer 1
Answer:

Answer:

a)(dB)/(dt) = (5\pi)/(4.2) \cdot \cos \left(2\pi\cdot (t)/(4.2) \right), b)(dB)/(dt)\approx 5.595

Step-by-step explanation:

a) The rate of change of the brightness of the Cepheid can be determined by deriving the function in time:

(dB)/(dt) = \left((2\pi)/(4.2) \right)\cdot 0.25\cdot \cos (2\pi\cdot (t)/(4.2))

(dB)/(dt) = (5\pi)/(4.2) \cdot \cos \left(2\pi\cdot (t)/(4.2) \right)

b) The rate of increase after one day is:

(dB)/(dt) = (5\pi)/(4.2) \cdot \left(2\pi \cdot (1)/(4.2) \right)

(dB)/(dt)\approx 5.595

Related Questions

A family has two cars. The first car has a fuel efficiency of 40 miles per gallon of gas and the second has a fuel efficiency of 20 miles per gallon of gas. During one particular week, the two cars went a combined total of 2200 miles, for a total gas consumption of 75 gallons. How many gallons were consumed by each of the two cars that week?
The owner of a grocery store wants to mix two kinds of candy together to make 15 lb that he can sell for $5.00 per lb. He wants to use chocolate candies that he sells for $7.00 per lb and sugar candies that he sells for $2.00 per lb. How many pounds of each should the owner use?_ pounds of chocolate candies_ pounds of sugar candies=OWO=
A trader borrowed 2500$ at a sumple interest at the end of 8 months he paid back $2500 find the rate​
Write the answer and solution ?
Course hero; suppose you are assigned the number 1, and the other students in your statistics class call out consecutive numbers until each person in the class has his or her own number. (a) explain how you could get a random sample of four students from your statistics class.

A given binomial distribution has 10 trials and probability of success p=1/3. Compute the standard deviation and explain your solution

Answers

Answer: 1.49

Step-by-step explanation:

We know that the standard deviation in a binomial distribution is given by :-

\sigma=√(n\cdot p\cdot (1-p)), where n is the number of trials and p is the probability of success.

Given : The number of trials : n=10

The probability of success = p=(1)/(3)

Then , the standard deviation will be :-

\sigma=\sqrt{10\cdot (1)/(3)\cdot (1-(1)/(3))}\n\n\Rightarrow\ \sigma=1.490711985\approx1.49

Find three consecutive numbers whose sum is 612

Answers

Answer: 203, 204, 205

Step-by-step explanation:

Consecutive numbers are numbers that are right next to each other. An example would be 1,2,3 or 67,68,69. Since this problem is asking for three consecutive numbers where the sum is equal to 612, we need to find the numbers that will be greater than 200.

Since we know that the sum is 612, we know that the numbers will be in the 2 hundreds. All we need to do is find 3 numbers that equal to 12.

3+4+5=12

Now that we have the three numbers, let's add 200.

203+204+205=612

Therefore, the three numbers are 203, 204, 205.

Answer: 203, 204, 205

Step-by-step explanation:

Let's say the number is x.

So the three consecutive numbers are: x, x + 1, and x+ 2

3x + 3 = 612

3x = 609

x = 203

Plug in x.

203, 204, and 205

Name the coefficients, the like terms, and constants.
8jk - 2 + 4kj - 3k - 9

Answers

2jk- 11-3k, I think that’s the answer

Please help me understand I am confused

Answers

9514 1404 393

Explanation:

a) The velocity curve is linearly increasing from 0 to 6 m/s over a period of 2 seconds, then linearly decreasing from 6 m/s to 0 over the same period. The acceleration is the rate of change of velocity, so for the first half of the motion the acceleration is a constant (6 m/s)/(2 s) = 3 m/s². Similarly, over the second half of the motion, the acceleration is a constant (-6 m/s)/(2 s) = -3 m/s².

The distance traveled is the integral of the velocity, so the linearly increasing velocity will cause the distance vs. time curve to have a parabolic shape. The shape will likewise be parabolic, but with decreasing slope, as the velocity ramps down to zero. Overall, the distance versus time curve will have an "S" shape.

The motion (position and velocity) will be continuous, but the acceleration will not be. There will be a significant "j.erk" at the 2-second mark where acceleration abruptly changes from increasing the velocity to braking (decreasing the velocity).

__

b) The attachment shows the (given) velocity curve in meters per second and its integral, the position curve, in meters.

The integral in the attached works nicely for machine evaluation. For hand evaluation, it is perhaps best written piecewise:

  s(t)=\begin{cases}\displaystyle\int_0^t{3x}\,dx\qquad\text{for $x\le2$}\n\n\displaystyle6+\int_2^t{(12-3x)}\,dx\qquad\text{for $2<x\le4$}\end{cases}

Limit of x^2-81/x+9
As x goes toward -9

Answers

Hello,

Use the factoration

a^2 - b^2 = (a - b)(a + b)

Then,

x^2 - 81 = x^2 - 9^2

x^2 - 9^2 = ( x - 9).(x + 9)

Then,

Lim (x^2- 81) /(x+9)

= Lim (x -9)(x+9)/(x+9)

Simplity x + 9

Lim (x -9)

Now replace x = -9

Lim ( -9 -9)

Lim -18 = -18
_______________

The second method without using factorization would be to calculate the limit by the hospital rule.

Lim f(x)/g(x) = lim f(x)'/g(x)'

Where,

f(x)' and g(x)' are the derivates.

Let f(x) = x^2 -81

f(x)' = 2x + 0
f(x)' = 2x

Let g(x) = x +9

g(x)' = 1 + 0
g(x)' = 1

Then the Lim stay:

Lim (x^2 -81)/(x+9) = Lim 2x /1

Now replace x = -9

Lim 2×-9 = Lim -18

= -18

Each prize for a carnival booth costs $0.32. How many prizes can you buy with $96?​

Answers

Answer: 96 divided by 0.32 is 300, so it's 300.
Other Questions
PLZ HELP ASAP CONGRUENCE
Find the value of 9 +y when y= 16.
Choose the correct simplification of 9x^2(4x + 2x^2 − 1)
Find the median of the data of 10, 15 , 6​