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2 years ago

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Select the correct word to describe the function. the equation y = startfraction 5 x cubed over 8 endfraction represents functio

n. the equation y = 6 + 0.25 ln x represents function. the equation y = startfraction 18 over 1 + 6 e superscript negative 0.25 x baseline endfraction represents function. the equation y = 5x3 + 8 represents function.

1 answer:

balandron [24]2 years ago

7 0

The functions and their descriptions are:

Polynomial: $y = \frac{5x^3}{8}$ and $y = 5x^3 + 8$
Logarithm: $y = 6 + 0.25\ln(x)$
Exponential $y = \frac{18}{1 + 6e^{-0.25x}}$

<h3>How to describe the functions?</h3>

The functions are given as:

$y = \frac{5x^3}{8}$

$y = 6 + 0.25\ln(x)$

$y = \frac{18}{1 + 6e^{-0.25x}}$

$y = 5x^3 + 8$

The functions would be described based on the type of function they represent.

Functions that use the "ln" keyword are logarithmic functions, while functions that use the "e" keyword are exponential functions.

Read more about functions at:

brainly.com/question/4025726

#SPJ1

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Solve the equation. <br><br> 2x+4+8x=24

Rasek [7]

2x + 4 + 8x = 24

2x + 8x = 24 - 4

10x = 20

x = 20 ÷ 10

x = 2

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Hope it helps!

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3 years ago

Read 2 more answers

A metal cylinder can with an open top and closed bottom is to have volume 4 cubic feet. Approximate the dimensions that require

Aleksandr-060686 [28]

Answer:

$r\approx 1.084\ feet$

$h\approx 1.084\ feet$

$\displaystyle A=11.07\ ft^2$

Step-by-step explanation:

<u>Optimizing With Derivatives </u>

The procedure to optimize a function (find its maximum or minimum) consists in :

Produce a function which depends on only one variable
Compute the first derivative and set it equal to 0
Find the values for the variable, called critical points
Compute the second derivative
Evaluate the second derivative in the critical points. If it results positive, the critical point is a minimum, if it's negative, the critical point is a maximum

We know a cylinder has a volume of 4 $ft^3$ . The volume of a cylinder is given by

$\displaystyle V=\pi r^2h$

Equating it to 4

$\displaystyle \pi r^2h=4$

Let's solve for h

$\displaystyle h=\frac{4}{\pi r^2}$

A cylinder with an open-top has only one circle as the shape of the lid and has a lateral area computed as a rectangle of height h and base equal to the length of a circle. Thus, the total area of the material to make the cylinder is

$\displaystyle A=\pi r^2+2\pi rh$

Replacing the formula of h

$\displaystyle A=\pi r^2+2\pi r \left (\frac{4}{\pi r^2}\right )$

Simplifying

$\displaystyle A=\pi r^2+\frac{8}{r}$

We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero

$\displaystyle A'=2\pi r-\frac{8}{r^2}=0$

Rearranging

$\displaystyle 2\pi r=\frac{8}{r^2}$

Solving for r

$\displaystyle r^3=\frac{4}{\pi }$

$\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet$

Computing h

$\displaystyle h=\frac{4}{\pi \ r^2}\approx 1.084\ feet$

We can see the height and the radius are of the same size. We check if the critical point is a maximum or a minimum by computing the second derivative

$\displaystyle A''=2\pi+\frac{16}{r^3}$

We can see it will be always positive regardless of the value of r (assumed positive too), so the critical point is a minimum.

The minimum area is

$\displaystyle A=\pi(1.084)^2+\frac{8}{1.084}$

$\boxed{ A=11.07\ ft^2}$

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2 years ago

At a sporting good store the price of a basketball at $650 during a sale this prices marked down 45% what is the sale price of a

Ne4ueva [31]

The multiplier for decreasing by 45% is 0.55
So you times the multiplier by the price to get your answer:

650 x 0.55 = $357.50

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3 years ago

The average rainfall for each week for the last 4 weeks was 7/12 inch. How much rain fell during the last 4 weeks?

kenny6666 [7]

7/12 times 4 is 2 1/3. So 2 1/3 inches each week

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3 years ago

Find the indicated limit, if it exists.

kondor19780726 [428]

Answer:

d) The limit does not exist

General Formulas and Concepts:

<u>Calculus</u>

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$
Left-Side Limit: $\displaystyle \lim_{x \to c^-} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Step-by-step explanation:

*Note:

In order for a limit to exist, the right-side and left-side limits must equal each other.

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle f(x) = \left\{\begin{array}{ccc}5 - x,\ x < 5\\8,\ x = 5\\x + 3,\ x > 5\end{array}$

<u>Step 2: Find Right-Side Limit</u>

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^+} 5 - x$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} 5 - x = 5 - 5 = 0$

<u>Step 3: Find Left-Side Limit</u>

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^-} x + 3$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} x + 3 = 5 + 3 = 8$

∴ Since $\displaystyle \lim_{x \to 5^+} f(x) \neq \lim_{x \to 5^-} f(x)$ , then $\displaystyle \lim_{x \to 5} f(x) = DNE$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

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2 years ago