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3 years ago

Is 3/8 less than 11/12

Mathematics

2 answers:

gizmo_the_mogwai [7]3 years ago

6 0

Yes it's less because3/8=8/32. 11/12=22/24

dezoksy [38]3 years ago

3 0

3/8 = 9/24

11/12 = 22/24

3/8 is less than 11/12

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Calculus, question 5 to 5a

Llana [10]

5. Let $x = \sin(\theta)$ . Note that we want this variable change to be reversible, so we tacitly assume 0 ≤ θ ≤ π/2. Then

$\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}$

and $dx = \cos(\theta) \, d\theta$ . So the integral transforms to

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \int \frac{\sin^3(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int \sin^3(\theta) \, d\theta$

Reduce the power by writing

$\sin^3(\theta) = \sin(\theta) \sin^2(\theta) = \sin(\theta) (1 - \cos^2(\theta))$

Now let $y = \cos(\theta)$ , so that $dy = -\sin(\theta) \, d\theta$ . Then

$\displaystyle \int \sin(\theta) (1-\cos^2(\theta)) \, d\theta = - \int (1-y^2) \, dy = -y + \frac13 y^3 + C$

Replace the variable to get the antiderivative back in terms of x and we have

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\cos(\theta) + \frac13 \cos^3(\theta) + C$

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2} + \frac13 \left(\sqrt{1-x^2}\right)^3 + C$

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\frac13 \sqrt{1-x^2} \left(3 - \left(\sqrt{1-x^2}\right)^2\right) + C$

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \boxed{-\frac13 \sqrt{1-x^2} (2+x^2) + C}$

6. Let $x = 3\tan(\theta)$ and $dx=3\sec^2(\theta)\,d\theta$ . It follows that

$\cos(\theta) = \dfrac1{\sec(\theta)} = \dfrac1{\sqrt{1+\tan^2(\theta)}} = \dfrac3{\sqrt{9+x^2}}$

since, like in the previous integral, under this reversible variable change we assume -π/2 < θ < π/2. Over this interval, sec(θ) is positive.

Now,

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \int \frac{27\tan^3(\theta)}{\sqrt{9+9\tan^2(\theta)}} 3\sec^2(\theta) \, d\theta = 27 \int \frac{\tan^3(\theta) \sec^2(\theta)}{\sqrt{1+\tan^2(\theta)}} \, d\theta$

The denominator reduces to

$\sqrt{1+\tan^2(\theta)} = \sqrt{\sec^2(\theta)} = |\sec(\theta)| = \sec(\theta)$

and so

$\displaystyle 27 \int \tan^3(\theta) \sec(\theta) \, d\theta = 27 \int \frac{\sin^3(\theta)}{\cos^4(\theta)} \, d\theta$

Rewrite sin³(θ) just like before,

$\displaystyle 27 \int \frac{\sin(\theta) (1-\cos^2(\theta))}{\cos^4(\theta)} \, d\theta$

and substitute $y=\cos(\theta)$ again to get

$\displaystyle -27 \int \frac{1-y^2}{y^4} \, dy = 27 \int \left(\frac1{y^2} - \frac1{y^4}\right) \, dy = 27 \left(\frac1{3y^3} - \frac1y\right) + C$

Put everything back in terms of x :

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac1{\cos^3(\theta)} - \frac3{\cos(\theta)}\right) + C$

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac{\left(\sqrt{9+x^2}\right)^3}{27} - \sqrt{9+x^2}\right) + C$

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \boxed{\frac13 \sqrt{9+x^2} (x^2 - 18) + C}$

2(b). For some constants a, b, c, and d, we have

$\dfrac1{x^2+x^4} = \dfrac1{x^2(1+x^2)} = \boxed{\dfrac ax + \dfrac b{x^2} + \dfrac{cx+d}{x^2+1}}$

3(a). For some constants a, b, and c,

$\dfrac{x^2+4}{x^3-3x^2+2x} = \dfrac{x^2+4}{x(x-1)(x-2)} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac c{x-2}}$

5(a). For some constants a-f,

$\dfrac{x^5+1}{(x^2-x)(x^4+2x^2+1)} = \dfrac{x^5+1}{x(x-1)(x+1)(x^2+1)^2} \\\\ = \dfrac{x^4 - x^3 + x^2 - x + 1}{x(x-1)(x^2+1)^2} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac{cx+d}{x^2+1} + \dfrac{ex+f}{(x^2+1)^2}}$

where we use the sum-of-5th-powers identity,

$a^5 + b^5 = (a+b) (a^4-a^3b+a^2b^2-ab^3+b^4)$

4 0

2 years ago

WILL GIVE BRAINLIEST!!

MA_775_DIABLO [31]

given a polynomial with factor (x - a)

Then x = a is a root of the polynomial and f(a) = 0

(1)

(x+ 2) is a factor , hence x = - 2 is a root

Evaluate the polynomial for x = - 2

f(- 2) = 5(- 2)³ + 8(- 2)² - 7(- 2) - 6 = - 40 + 32 + 14 - 6 = 0

Hence (x + 2 ) is a factor

(2)

If (x + 1) is a factor then x = - 1 is a root

5(- 1)³ + 8(-1)² - 7(- 1) - 6 = -5 + 8 + 7 - 6 = 4

Since f( - 1) ≠ 0

Then (x + 1) is not a factor

5 0

3 years ago

A forest ranger wants to install a 1,500 CFM (cubic feet per minute) fan. A ranger station consists of a rectangular prism and a

levacccp [35]

Sample response: <span>Yes, the fan will provide enough airflow. The volume of the ranger station is 720 cubic feet. The ranger station requires a 720 CFM fan. 1,500 CFM is greater than 720 CFM, so it will be enough airflow.</span>

6 0

3 years ago

Read 2 more answers

Write a decimal number<br>ral for 3/5

Jobisdone [24]

Answer:

0.6 or 0.60

Step-by-step explanation:

3/5 mulitply denominator to see how much it goes to 100 which is 20 so 5x20 which is 100 and u go to other number u get 60 so 60/100 or you can just divide 3 by 5 which is also 0.6

5 0

3 years ago

Read 2 more answers

Can y’all help me please

matrenka [14]

Answer:

w = 5

Step-by-step explanation:

The perimeter of a rectangle is given by

P = 2(l+w)

28 = 2(2w-1 + w)

Combine like terms

28 =2(3w-1)

Divide each side by 2

28/2 = 2/2(3w-1)

14 = 3w-1

Add 1 to each side

14+1 = 3w-1+1

15 = 3w

Divide each side by 3

15/3 =3w/3

5 =w

8 0

3 years ago