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wolverine [178]

4 years ago

13

A cylindrical pressure vessel has a height of 9 feet and a diameter of 6 feet. How much gas can the pressure vessel hold when fu

ll?
A. 54π
cubic feet

B. 81π
cubic feet

C. 108π
cubic feet

D. 324π
cubic feet

2 answers:

yawa3891 [41]4 years ago

7 0

Answer: B. 81π cubic feet

Step-by-step explanation:

The equation for cylinders V = π * r² * h

Plug in the numbers r = radius which is half of the diameter and h = height. Also leave out π because of the answer choices.

so... 3² * 9 = 81

81π

natulia [17]4 years ago

5 0

Hi There ^^

Answer :- ( B ) 81 π cubic feet

Explaination :-

$We \: \: are \: \: given \: \: that \\ \\ \: \\ \: Height \: = \: 9 \: feet \\ \\ \\Diameter \: = \: 6 \: feet \\ \\ \\radius = 3 feet \\ \\ \\ Volume \: \: = \pi {r}^{2} h \\ \\ = \pi \times 3 \times 3 \times 9 \\ \\ = 81\pi \: {feet}^{3}$
Thanks

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Si un gramo de chocolate vale$38.25¿cuánto costarán 750 gramos?

Anna007 [38]

Answer:

28687.5

Step-by-step explanation:

Un gramo equivale a $38,25, por lo que si tenemos 750 de esos gramos, tendríamos que multiplicarlos para obtener el costo total.

3 0

2 years ago

The trees at a national park have been increasing in numbers. There were 1,000 trees in the first year that the park started tra

ivanzaharov [21]

The sequence forms a Geometric sequence as the rule to obtain the value for the next term is by ratio

Term 1: 1000
Term 2: 200
Term 3: 40

From term 1 to term 2, there's a decrease by $\frac{1}{5}$
From term 2 to term 3, there's a decrease also by $\frac{1}{5}$

The rule to find the $n^{th}$ term in a sequence is
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So, the formula for the sequence in question is
$n_{th}$ term = $1000( \frac{1}{5} ^{n-1} )$

The sequence is a divergent one. We can always find the value of the next term by dividing the previous term by 5 and if we do that, the value of the next term will get closer to 'zero' but never actually equal to zero.

We can find a partial sum of the sequence using the formula
$S_{∞} = \frac{a}{1-r}$ for -1<r<1
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6 0

3 years ago

Student Produced Response - Calculator

faltersainse [42]

Answer:

$\dfrac{a}{b}=0.75$ .

Step-by-step explanation:

If two equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ has infinitely many solutions, then

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

The given equations are

$ax+by=10$

$3x+4y=20$

It is given that the above system of equations has infinitely many solutions. So,

$\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{10}{20}$

$\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{1}{2}$

Now,

$\dfrac{a}{3}=\dfrac{1}{2}$ and $\dfrac{b}{4}=\dfrac{1}{2}$

$a=\dfrac{3}{2}$ and $b=\dfrac{4}{2}$

$a=1.5$ and $b=2$

So, a=1.5 and b=2.

Now,

$\dfrac{a}{b}=\dfrac{1.5}{2}=0.75$

Therefore, $\dfrac{a}{b}=0.75$ .

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4 years ago

The franklins have a home loan with an interest rate of 4 1/4%. write the percent as a fraction in simplest form.

alex41 [277]

4 1/4 % =
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8 0

3 years ago

Read 2 more answers

The third-degree Taylor polynomial about x = 0 of In(1 - x) is

gizmo_the_mogwai [7]

Answer:

$\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Algebra I</u>

Functions
Function Notation

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

MacLaurin/Taylor Polynomials

Approximating Transcendental and Elementary functions
MacLaurin Polynomial: $\displaystyle P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... + \frac{f^{(n)}(0)}{n!}x^n$
Taylor Polynomial: $\displaystyle P_n(x) = \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n$

Step-by-step explanation:

*Note: I will not be showing the work for derivatives as it is relatively straightforward. If you request for me to show that portion, please leave a comment so I can add it. I will also not show work for elementary calculations.

<u />

<u>Step 1: Define</u>

<em>Identify</em>

f(x) = ln(1 - x)

Center: x = 0

<em>n</em> = 3

<u>Step 2: Differentiate</u>

[Function] 1st Derivative: $\displaystyle f'(x) = \frac{1}{x - 1}$
[Function] 2nd Derivative: $\displaystyle f''(x) = \frac{-1}{(x - 1)^2}$
[Function] 3rd Derivative: $\displaystyle f'''(x) = \frac{2}{(x - 1)^3}$

<u>Step 3: Evaluate Functions</u>

Substitute in center <em>x</em> [Function]: $\displaystyle f(0) = ln(1 - 0)$
Simplify: $\displaystyle f(0) = 0$
Substitute in center <em>x</em> [1st Derivative]: $\displaystyle f'(0) = \frac{1}{0 - 1}$
Simplify: $\displaystyle f'(0) = -1$
Substitute in center <em>x</em> [2nd Derivative]: $\displaystyle f''(0) = \frac{-1}{(0 - 1)^2}$
Simplify: $\displaystyle f''(0) = -1$
Substitute in center <em>x</em> [3rd Derivative]: $\displaystyle f'''(0) = \frac{2}{(0 - 1)^3}$
Simplify: $\displaystyle f'''(0) = -2$

<u>Step 4: Write Taylor Polynomial</u>

Substitute in derivative function values [MacLaurin Polynomial]: $\displaystyle P_3(x) = \frac{0}{0!} + \frac{-1}{1!}x + \frac{-1}{2!}x^2 + \frac{-2}{3!}x^3$
Simplify: $\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

Topic: AP Calculus BC (Calculus I/II)

Unit: Taylor Polynomials and Approximations

Book: College Calculus 10e

5 0

3 years ago