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sashaice [31]
1 year ago
9

I need help with this practice problem It is from my ACT prep guide It asks to graph the function, if you can, use Desmos to com

plete

Mathematics
1 answer:
Vaselesa [24]1 year ago
6 0

Given the following function:

\text{ f(x) = cot(x + }\frac{\pi}{6})

Use the form a·cot(bx - c) to find the variables used to find the amplitude, period, phase shift and vertical shift.

We get,

a = 1

b = 1

c = -π/6

d = 0

Since the graph of the Cotangent function does not have a maximum or minimum value, there can be no value for the amplitude.

Amplitude: None

For the Period:

\text{ Period = }\frac{\text{ }\pi}{\text{ b}}\text{ = }\frac{\pi}{1}\text{ = }\pi

Therefore, the Period is π.

For the Phase Shift:

\text{ Phase Shift = }\frac{\text{ c}}{\text{ b}}\text{ = }\frac{-\frac{\pi}{6}}{1}\text{ = -}\frac{\pi}{6}

Therefore, Phase Shift is -π/6 → π/6 (To the left).

For the Vertical Shift:

Vertical Shift = 0

Plotting the function will be:

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D" id="TexFormula1" title="\frac{1}{9}" alt="\frac{1}{9}" align="absmiddle
Jobisdone [24]

\dfrac{1}{9} - \dfrac{5}{6}

  • Set up

\dfrac{2}{18} - \dfrac{15}{18}

  • Get all fractions to a common denominator which will allow you to subtract them

\dfrac{2 - 15}{18}

  • Subtract numerators

\boxed{- \dfrac{13}{18}}

  • Simplify
4 0
3 years ago
Adding cheese to a McDonald's Quarter Pounder increases the fat content from 21 g to 30 g. What percent does the fat content inc
solong [7]

Answer: 42.9%

Step-by-step explanation:

The percentage increase of the fat content will be calculated as:

= Fat increase / Previous fat content × 100

= (30 - 21) / 21 × 100.

= 9/21 × 100

= 42.9%

5 0
3 years ago
In the​ 1970s, due to world​ events, there was a gasoline shortage in the United States. There were often long lines of cars wai
barxatty [35]

Answer:

Part A: The average car length is 9.1 feet to the nearest tenth foot

Part B: The line would be 9100 feet to contained 1000 cars

Step-by-step explanation:

* Lets explain how to solve the problem

# Part A:

- There were 62 cars in a line that stretched 567 feet

- The cars are lined up bumper-to-bumper

- That means there is no empty spaces between the cars

* To find the average length of the car we will divide the length of

  the line by the numbers of the cars

∵ The average car length = length of the line/number of cars

∵ The length of the line is 567 feet

∵ The numbers of the cars is 62 cars

∴ The average car length = 567/62 = 9.1 feet

* The average car length is 9.1 feet to the nearest tenth foot

# Part B:

- There are 1000 cars

- We need to find the length of line which contained the cars

∵ The average car length = length of the line/number of cars

∵ The average of car length is 9.1 feet

∵ The number of the cars is 1000 cars

∴ 9.1 = length of the line/1000

- Multiply both sides by 1000

∴ The length of the line = 9.1 × 1000 = 9100 feet

∴ The line would be 9100 feet to contained 1000 cars

7 0
4 years ago
The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability th
maksim [4K]

Answer:

The probability that there are 3 or less errors in 100 pages is 0.648.        

Step-by-step explanation:

In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.

For the given Poisson distribution the mean is p = 0.03 errors per page.

We have to find the probability that there are three or less errors in n = 100 pages.

Let us denote the number of errors in the book by the variable x.

Since there are on an average 0.03 errors per page we can say that

the expected value is, \lambda = E(x)

                                       = n × p

                                       = 100 × 0.03

                                       = 3

Therefore the we find the probability that there are 3 or less errors on the page as

     P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)

                 

   Using the formula for Poisson distribution for P(x = X ) = \frac{e^{-\lambda}\lambda^X}{X!}

Therefore P( X ≤ 3) = \frac{e^{-3} 3^0}{0!} + \frac{e^{-3} 3^1}{1!} + \frac{e^{-3} 3^2}{2!} + \frac{e^{-3} 3^3}{3!}

                                 = 0.05 + 0.15 + 0.224 + 0.224

                                 = 0.648

 The probability that there are 3 or less errors in 100 pages is 0.648.                                

7 0
3 years ago
Read 2 more answers
PLEASSEE HELPP!!
svp [43]
A dilation with scale factor 3 gives the effect of stretching the line AB three times longer, as shown in the diagram below

The slope will stay the same as dilation does not change the direction of the line

If point O lies on AB and is the center of dilation, then the point O must also lie on A'B'

Answer:
the slope of A'B' = 3
A'B' passes through point O


4 0
3 years ago
Read 2 more answers
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