Which equation can be used to solve for x in the following diagram.

Let the region R be the area enclosed by the function f(x)=ln(x) and g(x)= 1/2x-2. Find the volume of the solid generated when t

Neporo4naja [7]

Answer:

$V=61.66$

Step-by-step explanation:

This problem can be solved by using the expression for the Volume of a solid with the washer method

$V=\pi \int \limit_a^b[R(x)^2-r(x)^2]dx$

where R and r are the functions f and g respectively (f for the upper bound of the region and r for the lower bound).

Before we have to compute the limits of the integral. We can do that by taking f=g, that is

$f(x)=g(x)\\ln(x)=\frac{1}{2}x-2$

there are two point of intersection (that have been calculated with a software program as Wolfram alpha, because there is no way to solve analiticaly)

x1=0.14

x2=8.21

and because the revolution is around y=-5 we have

$R=ln(x)-(-5)\\r=\frac{1}{2}x-2-(-5)\\$

and by replacing in the integral we have

$V=\pi \int \limit_{x1}^{x2}[(lnx+5)^2-(\frac{1}{2}x+3)^2]dx\\$

$V=\pi [28x+\frac{1}{x}+xln^2x-12xlnx-6lnx]$

and by evaluating in the limits we have

$V=61.66$

Hope this helps

regards

3 0

3 years ago

The mean heigh of American Males is 69.5 inches. The height of the 43 male presidents have a mean 70.78 inches and a standard de

Arisa [49]

Answer:

There is sufficient statistical evidence to suggest that the US presidents are taller than the average American male

Step-by-step explanation:

The given parameters are;

The mean height of American Males, μ = 69.5 inches

The mean height of 43 male presidents, $\overline x$ = 70.78

The standard deviation, s = 2.77 inches

The number of male presidents = 43

The null hypothesis; H₀ μ = $\overline x$

The alternative hypothesis; Hₐ μ ≠ $\overline x$

The significance level, α = 0.05

The t test for the sample with unknown population standard deviation is given as follows;

$t=\dfrac{\bar{x}-\mu }{\dfrac{s }{\sqrt{n}}}$

Therefore, we have;

$t=\dfrac{70.78-69.5 }{\dfrac{2.77 }{\sqrt{43}}} \approx 3.0302$

The degrees of freedom, df = n - 1 = 43 - 1 = 42

The critical 't' value = 2.0181

Therefore, given that the t test value is larger than the critical-t, we reject the null hypothesis, therefore, there is sufficient statistical evidence to suggest that the US presidents are taller than the average American male

4 0

3 years ago

standard automobile license plates in a country display 3 numbers, followed by 2 letters, followed by 2 numbers. how many differ

RoseWind [281]

The system of standard automotive license plates allows for a variety of standard plates is 39917124 ways.

<h3>Define the term permutation?</h3>

The number of potential arrangements for a given set is assessed mathematical terms, but this procedure is referred to as permutation.
The arrangement's order is essential when using permutations.

For the stated data;

Standard car license plates in a nation typically have 3 digits, 2 letters, and 2 numbers.

There are total 9 digits (0 - 9).

There are total 26 letter (A - Z)

Repetitions of letters and numbers given allowed.

Thus, numbers and the letters ca be arranged as;

1st place = 9 ways.

2nd place = 9 ways.

3nd place = 9 ways.

4th place = 26 ways.

6th place = 26 ways.

7th place = 9 ways.

8th place = 9 ways.

Total ways = 9×9×9×26×26×9×9

Total ways = 39917124

Thus, the system of standard automotive license plates allows for a variety of standard plates is 39917124 ways.

To know more about the permutation, here

brainly.com/question/1216161

#SPJ4

5 0

1 year ago

NEED HELP ASAP!!

Solnce55 [7]

Answer:

Only Cory is correct

Step-by-step explanation:

The gravitational pull of the Earth on a person or object is given by Newton's law of gravitation as follows;

$F =G\times \dfrac{M \cdot m}{r^{2}}$

Where;

G = The universal gravitational constant

M = The mass of one object

m = The mass of the other object

r = The distance between the centers of the two objects

For the gravitational pull of the Earth on a person, when the person is standing on the Earth's surface, r = R = The radius of the Earth ≈ 6,371 km

Therefore, for an astronaut in the international Space Station, r = 6,800 km

The ratio of the gravitational pull on the surface of the Earth, F₁, and the gravitational pull on an astronaut at the international space station, F₂, is therefore given as follows;

$\dfrac{F_1}{F_2} = \dfrac{ \dfrac{M \cdot m}{R^{2}}}{\dfrac{M \cdot m}{r^{2}}} = \dfrac{r^2}{R^2} = \dfrac{(6,800 \ km)^2}{(6,371 \ km)^2} \approx 1.14$

∴ F₁ ≈ 1.14 × F₂

F₂ ≈ 0.8778 × F₁

Therefore, the gravitational pull on the astronaut by virtue of the distance from the center of the Earth, F₂ is approximately 88% of the gravitational pull on a person of similar mass on Earth

However, the International Space Station is moving in its orbit around the Earth at an orbiting speed enough to prevent the Space Station from falling to the Earth such that the Space Station falls around the Earth because of the curved shape of the gravitational attraction, such that the astronaut are constantly falling (similar to falling from height) and appear not to experience gravity

Therefore, Cory is correct, the astronauts in the International Space Station, 6,800 km from the Earth's center, are not too far to experience gravity.

6 0

3 years ago

In a salad recipe, the ratio of carrots to cucumbers must remain constant. The table below shows some possible combinations of c

skelet666 [1.2K]

I think the answer is B

4 0

3 years ago