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Vinil7 [7]
3 years ago
11

Find the volume of the cylinder. Use pi = 3.14* the height of 6 inches and a radius 4.5in​

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
8 0

Answer: 381.51 cubic inches

Step-by-step explanation:

1. Use the formula: V = πr^{2}h, or in this case, 3.14(4.5^{2})(6)

2. Simplify to 3.14(20.25)(6)

3. Simply multiply all three numbers using a calculator, or one by one if that's the preferred option.

4. You get: 381.51 cubic inches.

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Step-by-step explanation:

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Step-by-step explanation:

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Inequalities Word Problem:
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Answer:

37.5 hours.

Step-by-step explanation:

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3 0
2 years ago
If a and b are positive numbers, find the maximum value of f(x) = x^a(2 − x)^b on the interval 0 ≤ x ≤ 2.
Ad libitum [116K]

Answer:

The maximum value of f(x) occurs at:

\displaystyle x = \frac{2a}{a+b}

And is given by:

\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

Step-by-step explanation:

Answer:

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b >0

And we want to find the maximum value of f(x) on the interval [0, 2].

First, let's evaluate the endpoints of the interval:

\displaystyle f(0) = (0)^a(2-(0))^b = 0

And:

\displaystyle f(2) = (2)^a(2-(2))^b = 0

Recall that extrema occurs at a function's critical points. The critical points of a function at the points where its derivative is either zero or undefined. Thus, find the derivative of the function:

\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right]

By the Product Rule:

\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}

Set the derivative equal to zero and solve for <em>x: </em>

\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right]

By the Zero Product Property:

\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0

The solutions to the first equation are <em>x</em> = 0 and <em>x</em> = 2.

First, for the second equation, note that it is undefined when <em>x</em> = 0 and <em>x</em> = 2.

To solve for <em>x</em>, we can multiply both sides by the denominators.

\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))

Simplify:

\displaystyle a(2-x) - b(x) = 0

And solve for <em>x: </em>

\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\  \frac{2a}{a+b} &= x  \end{aligned}

So, our critical points are:

\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}

We already know that f(0) = f(2) = 0.

For the third point, we can see that:

\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(2- \frac{2a}{a+b}\right)^b

This can be simplified to:

\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

Since <em>a</em> and <em>b</em> > 0, both factors must be positive. Thus, f(2a / (a + b)) > 0. So, this must be the maximum value.

To confirm that this is indeed a maximum, we can select values to test. Let <em>a</em> = 2 and <em>b</em> = 3. Then:

\displaystyle f'(x) = x^2(2-x)^3\left(\frac{2}{x} - \frac{3}{2-x}\right)

The critical point will be at:

\displaystyle x= \frac{2(2)}{(2)+(3)} = \frac{4}{5}=0.8

Testing <em>x</em> = 0.5 and <em>x</em> = 1 yields that:

\displaystyle f'(0.5) >0\text{ and } f'(1)

Since the derivative is positive and then negative, we can conclude that the point is indeed a maximum.

Therefore, the maximum value of f(x) occurs at:

\displaystyle x = \frac{2a}{a+b}

And is given by:

\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b

5 0
3 years ago
Order the numbers 21 , 40, and 42 from least to greatest based on how many numbers are in their prime factorization.
lora16 [44]

Answer:

21,40,42

Step-by-step explanation:

That's it.Wish it's correct.:D

6 0
3 years ago
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