1. Determine the intervals on which each function is increasing or
decreasing
(a) f(x) = x^3 - 11/2x^2 - 4x

Answers

Answer 1

Answer:

[tex]\text{$f(x)$ is increasing for $(-\infty, -\frac{1}{3})$ and $(4, \infty)$} \\ \text{$f(x)$ is decreasing for $(-\frac{1}{3}, 4)$}[/tex]

Step-by-step explanation:

We have the function:

[tex]f(x)=x^3-\frac{11}{2}x^2-4x[/tex]

And we want to determine the intervals for which the function is increasing or decreasing.

We will need to first find the critical points of the function. Note that our original function is continuous across the entire x-axis.

To find the critical points, we need to find the first derivative and then solve for the x. So:

[tex]f^\prime(x)=\frac{d}{dx}[x^3-\frac{11}{2}x^2-4x][/tex]

Differentiate:

[tex]f^\prime(x)=(3x^2)-\frac{11}{2}(2x)-(4) \\ f^\prime(x)=3x^2-11x-4[/tex]

Set the derivative equal to 0 and solve for x:

[tex]0=3x^2-11x-4[/tex]

Factor:

[tex]0=3x^2-12x+x-4 \\ 0=3x(x-4)+(x-4) \\ 0=(3x+1)(x-4)[/tex]

Zero Product Property:

[tex]3x+1=0\text{ or } x-4=0 \\ x=-\frac{1}{3}\text{ or } x=4[/tex]

Therefore, our critical points are -1/3 and 4.

We can thus sketch the following number line:

<————-(-1/3)——————————————(4)—————>

Now, let’s test values for the three intervals: less than -1/3, between -1/3 and 4, and greater than 4.

For less then -1/3, we can use -1. So, substitute -1 for x for our first derivative and see which we get:

[tex]f^\prime(-1)=3(-1)^2-11(-1)-4=3+11-4=10>0[/tex]

The result is positive.

Therefore, for all numbers less than -1/3,f(x) is increasing (since its derivative is positive).

For between -1/3 and 4, we can use 0. Substitute 0 for our derivative:

[tex]f^\prime(0)=3(0)^2-11(0)-4=-4<0[/tex]

The result is negative.

Therefore, for all numbers between -1/3 and 4, f(x) is decreasing (since its derivative is negative).

And finally, for greater than 4, we can use 5:

[tex]f^\prime(5)=3(5)^2-11(5)-4=16>0[/tex]

The result is positive.

Therefore, for all numbers greater than 4, f(x) is increasing (since its derivative is positive.

Therefore:

[tex]\text{$f(x)$ is increasing for $x<-\frac{1}{3}$ and $x>4$}\\ \text{ $f(x)$ is decreasing for $-\frac{1}{3}<x<4$}[/tex]

In interval notation:

[tex]\text{$f(x)$ is increasing for $(-\infty, -\frac{1}{3})$ and $(4, \infty)$} \\ \text{$f(x)$ is decreasing for $(-\frac{1}{3}, 4)$}[/tex]


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3) The measure of the angle which is equal to its supplement is

Answers

Answer:

Step-by-step explanation:

90 degrees

x + y = 180

x = y

x + x = 180

2x = 180

x = 180/2

x = 90

Both the angle and its supplement = 90

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Answers

Answer:

7/3

Step-by-step explanation:

The slope is found by dividing the difference in the y-coordinates by the difference in the x-coordinates.

Here, the difference in the y-coordinates is: 4 - (-3) = 4 + 3 = 7. The difference in the x-coordinates is: 3 - 0 = 3.

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The slope you are looking for is 7.

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y = 4x + 2
y = [?]x-4
Enter the number that belongs in the green box.

Answers

Answer:

4

Step-by-step explanation:

Parallel lines have the same slope and since the slope of the first line is 4 the second lines slope is also 4

Answer:

Step-by-step explanation:

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Answers

Answer:

[tex]\frac{1}{2}[/tex]  which agrees with answer B

Step-by-step explanation:

First write the equation that represents this type of variation:

[tex]y=\frac{5}{x}[/tex]

then we need to solve for "x" when y  = 10 as shown below:

[tex]y=\frac{5}{x} \\10=\frac{5}{x} \\10*x=5\\x=\frac{5}{10} \\x=\frac{1}{2}[/tex]

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Answers

y intercept is 5.5
x intercept is -7.5

Answer:

X intercept is the point where the line meets the x. y intercept is the point where the line meets the y.  So, the x intercept is -7.5 and the y intercept is 5.5.

 

Consider the limaçon with equation r = 3 + 4cos(θ). How does the quotient of a and b relate to the existence of an inner loop?

Answers

A polar graph that is a limacon has a formula similar to [tex]r=a+bcos\theta[/tex]

Option B is correct.

A  limacon has a formula similar to [tex]r=a+bcos\theta[/tex]

Case 1 .  If  a < b  or  [tex]\frac{b}{a}>1[/tex]

Then the curve is limacon with an inner loop.

Case 2.  If  a>b   or   [tex]\frac{b}{a}<1[/tex]

Then the limacon does not have an inner loop.

Here, given that,  [tex]r=3+4cos\theta[/tex]

It is observed that,  a < b  or [tex]\frac{b}{a}>1[/tex]

Therefore, the curve is limacon with an inner loop.

Hence, option B is correct.

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

-39.75 = 7.5m

Divide both sides by 7.5 to get m alone

-39.75/7.5 = 7.5m/7.5

-5.3 = m

Hope this helps:)

Answer:

m= -5.3

Step-by-step explanation:

-39.75=7.5m

(divide both sides by 7.5)

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

104 divided by 8 is 13

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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If f(x)=2x+4 then the inverse is x=(2/7)f(x)+4
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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

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The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.

Required:
a. A random sample of six women is selected. What is the probability that the sample mean is greater than 63 inches?
b. What is the probability that a randomly selected woman is taller than 66 inches?
c. What is the probability that the mean height of a random sample of 100 women is greater than 66 inches?

Answers

Answer:

(a) 0.2061

(b) 0.2514

(c) 0

Step-by-step explanation:

Let X denote the heights of women in the USA.

It is provided that X follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

[tex]P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z<0.82)\\\\=0.20611\\\\\approx 0.2061[/tex]

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

[tex]P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z<0.67)\\\\=1-0.74857\\\\=0.25143\\\\\approx 0.2514[/tex]

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

[tex]P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0[/tex]

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

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