A plane is two-dimensional.

oksano4ka [1.4K]

3 × 1/10 + 6 × 1/100 + 6 × 1/1000
I'm pretty sure that's it but I'm not good at math

4 0

3 years ago

find the volume of the solid formed by revolving the region bounded by the graphs of y = 4x - x^2 and f(x) = x^2 from [0,2] abou

Neko [114]

Answer:

$v = \frac{32\pi}{3}$

or

$v=33.52$

Step-by-step explanation:

Given

$f(x) = 4x - x^2$

$g(x) = x^2$

$[a,b] = [0,2]$

Required

The volume of the solid formed

Rotating about the x-axis.

Using the washer method to calculate the volume, we have:

$\int dv = \int\limit^b_a \pi(f(x)^2 - g(x)^2) dx$

Integrate

$v = \int\limit^b_a \pi(f(x)^2 - g(x)^2)\ dx$

$v = \pi \int\limit^b_a (f(x)^2 - g(x)^2)\ dx$

Substitute values for a, b, f(x) and g(x)

$v = \pi \int\limit^2_0 ((4x - x^2)^2 - (x^2)^2)\ dx$

Evaluate the exponents

$v = \pi \int\limit^2_0 (16x^2 - 4x^3 - 4x^3 + x^4 - x^4)\ dx$

Simplify like terms

$v = \pi \int\limit^2_0 (16x^2 - 8x^3 )\ dx$

Factor out 8

$v = 8\pi \int\limit^2_0 (2x^2 - x^3 )\ dx$

Integrate

$v = 8\pi [ \frac{2x^{2+1}}{2+1} - \frac{x^{3+1}}{3+1} ]|\limit^2_0$

$v = 8\pi [ \frac{2x^{3}}{3} - \frac{x^{4}}{4} ]|\limit^2_0$

Substitute 2 and 0 for x, respectively

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ \frac{2*0^{3}}{3} - \frac{0^{4}}{4} ])$

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ 0 - 0])$

$v = 8\pi [ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ]$

$v = 8\pi [ \frac{16}{3} - \frac{16}{4} ]$

Take LCM

$v = 8\pi [ \frac{16*4- 16 * 3}{12}]$

$v = 8\pi [ \frac{64- 48}{12}]$

$v = 8\pi * \frac{16}{12}$

Simplify

$v = 8\pi * \frac{4}{3}$

$v = \frac{32\pi}{3}$

or

$v=\frac{32}{3} * \frac{22}{7}$

$v=\frac{32*22}{3*7}$

$v=\frac{704}{21}$

$v=33.52$

8 0

3 years ago

Can someone help me with #6? (area of circles)

Vlad1618 [11]

Qué es?

A.Es una grapadora. B.Es una puerta. C.Es un mapa. D.Es un cartel.

7 0

3 years ago

You measure 22 textbooks' weights, and find they have a mean weight of 64 ounces. Assume the population standard deviation is 5.

lara31 [8.8K]

Answer:

Step-by-step explanation:

We want to determine a 90% confidence interval for the true population mean textbook weight.

Number of sample, n = 22

Mean, u = 64 ounces

Standard deviation, s = 5.1 ounces

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

64 ± 1.645 × 5.1/√22

= 64 ± 1.645 × 1.087

= 64 ± 1.788

The lower end of the confidence interval is 64 - 1.788 = 62.21 ounces

The upper end of the confidence interval is 64 + 1.788 = 65.79 ounces

Therefore, with 90% confidence interval, the true population mean textbook weight is between 62.21 ounces and 65.79 ounces

3 0

3 years ago

Show your work for this one

Klio2033 [76]

Step-by-step explanation:

multipy the two sides by 2

n+4= -16

move the constant to the right change its sign

n= -16 - 4

calculate the difference

N= - 20

4 0

3 years ago

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