All categories

2 years ago

A Cepheid star is a type of variable star, which means that its brightness is not constant.

1 answer:

7 0

A function assigns the values. The absolute magnitude of a star that has a period of 62 days is -6.3328.

<h3>What is a Function?</h3>

A function assigns the value of each element of one set to the other specific element of another set.

The relationship between the brightness of a Cepheid star and its period, or length of its pulse, is given by

M = − 2.78 (log P) − 1.35,

Given the absolute magnitude of a star that has a period of 62 days is,

M = − 2.78 (log 62) − 1.35

M = -6.3328

Hence, the absolute magnitude of a star that has a period of 62 days is -6.3328.

Learn more about Function:

brainly.com/question/5245372

#SPJ1

You might be interested in

What is the solution to this equation?

slamgirl [31]

Answer: No solution.

Step-by-step explanation:

1. The lines are parallel.

2. If you solve the system of equations you would end up with 0x, so you cannot solve for x.

4 0

3 years ago

The price of an item has been reduced by 40%. The original price was $60. What is the price of the item now?

erik [133]

To reduce the item by 40%, you first need to find the 40% of 60. Multiply 0.4*60 and you will get 24. That is 40%. Now you need to subtract that from the original. 60-24 is 36 dollars and that is your answer.

4 0

4 years ago

What is the missing term in the factorization?

jasenka [17]

Answer:

the missing term is 5

Step-by-step explanation:

Hi there !

12x² - 75 =

= 3(4x² - 25)

use formula a² - b² = (a - b)(a + b)

= 3(2x - 5)(2x + 5)

= 3(2x + 5)(2x - 5)

the missing term is 5

Good luck !

4 0

3 years ago

Let f be the function defined by f(x) = e^(x) cos x.

Pavel [41]

(a)

The average rate of change of f on the interval 0 ≤ x ≤ π is

$\displaystyle
f_{avg\Delta} = \frac{f(\pi) - f(0)}{\pi - 0} =\frac{-e^\pi-1}{\pi}$

____________

(b)

$f(x) = e^{x} cos x \implies f'(x) = e^x \cos(x) - e^x \sin(x) \implies \\ \\
f'\left(\frac{3\pi}{2} \right) = e^{3\pi/2} \cos(3\pi/2) - e^{3\pi/2} \sin(3\pi/2) \\ \\
f'\left(\frac{3\pi}{2} \right) = 0 - e^{3\pi/2} (-1) = e^{3\pi/2}$

The slope of the tangent line is $e^{3\pi/2}$ .

____________

(c)

The absolute minimum value of f occurs at a critical point where f'(x) = 0 or at endpoints.

Solving f'(x) = 0

$f'(x) = e^x \cos(x) - e^x \sin(x) \\ \\
0 = e^x \big( \cos(x) - \sin(x)\big)$

Use zero factor property to solve.

$e^x \ \textgreater \ 0\forall x \in \mathbb{R}$ so that factor will not generate solutions.
Set cos(x) - sin(x) = 0

$\cos (x) - \sin (x) = 0 \\
\cos(x) = \sin(x)$

cos(x) = 0 when x = π/2, 3π/2, but x = π/2. 3π/2 are not solutions to the equation. Therefore, we are justified in dividing both sides by cos(x) to make tan(x):

$\displaystyle\cos(x) = \sin(x) \implies 0 = \frac{\sin (x)}{\cos(x)} \implies 0 = \tan(x) \implies \\ \\
x = \pi/4,\ 5\pi/4\ \forall\ x \in [0, 2\pi]$

We check the values of f at the end points and these two critical numbers.

$f(0) = e^1 \cos(0) = 1$

$\displaystyle f(\pi/4) = e^{\pi/4} \cos(\pi/4) = e^{\pi/4} \frac{\sqrt{2}}{2}$

$\displaystyle f(5\pi/4) = e^{5\pi/4} \cos(5\pi/4) = e^{5\pi/4} \frac{-\sqrt{2}}{2} = -e^{\pi/4} \frac{\sqrt{2}}{2}$

$f(2\pi) = e^{2\pi} \cos(2\pi) = e^{2\pi}$

There is only one negative number.
The absolute minimum value of f on the interval 0 ≤ x ≤ 2π is
$-e^{5\pi/4} \sqrt{2}/2$

____________

(d)

The function f is a continuous function as it is a product of two continuous functions. Therefore, $\lim_{x \to \pi/2} f(x) = f(\pi/2) = e^{\pi/2} \cos(\pi/2) = 0$

g is a differentiable function; therefore, it is a continuous function, which tells us $\lim_{x \to \pi/2} g(x) = g(\pi/2) = 0$ .

When we observe the limit $\displaystyle \lim_{x \to \pi/2} \frac{f(x)}{g(x)}$ , the numerator and denominator both approach zero. Thus we use L'Hospital's rule to evaluate the limit.

$\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \lim_{x \to \pi/2} \frac{f'(x)}{g'(x)} = \frac{f'(\pi/2)}{g'(\pi/2)}$

$f'(\pi/2) = e^{\pi/2} \big( \cos(\pi/2) - \sin(\pi/2)\big) = -e^{\pi/2} \\ \\
g'(\pi/2) = 2$

thus

$\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \frac{-e^{\pi/2}}{2}$

3 0

3 years ago

Alaine has 1 gallon of paint. She is going to pour it into a paint tray that measures 10 inches wide, 12 inches long, and 5 cm d

Damm [24]

If Alaine has 1 gallon of paint, and she pours it into a paint tray that measures 10"w, 12"L, and 3cm deep, the paint tray will not fill the tray by 5.22 in3.

6 0

4 years ago