Part 3: Write the equation of and graph an ellipse.
Given the foci and vertices of an ellipse, complete the following steps.
Vertices at (4, -6 +-9)
Foci at (4, -6 +-5sqrt2)

a) What is the center of the ellipse? Explain how to determine this from the given points.

b) Write the standard equation of the ellipse. (3 points

c) What are the lengths of the major and minor axes? (2 points)
Major Axis:
Minor Axis:

d) Sketch a graph of the ellipse and label the center, foci, and the major and minor axes. (4 points)

Part 3: Write The Equation Of And Graph An Ellipse.Given The Foci And Vertices Of An Ellipse, Complete

Answers

Answer 1

Answer:

Step-by-step explanation:

The center is halfway between vertices, at (4, -6).

It is also halfway between foci.

:::::

The vertices are vertically aligned, so the parabola is vertical.

General equation for a vertical ellipse:

 (y-k)²/a² + (x-h)²/b² = 1

with

 center (h,k)

 vertices (h,k±a)

 co-vertices (h±b,k)

 foci (h,k±c), c² = a²-b²

Apply your data and solve for h, k, a, and b.

center (h,k) = (4, -6)

h = 4

k = -6

vertices (4,-6±a) = (4,-6±9)

a = 9

foci (4,-6±c) = (4,-6±5√2)

b² = a² - c² = 9² - (5√2)² = 31

b = √31

The equation becomes

 (y+6)²/81 + (x-4)²/31 = 1

:::::

length of major axis = 2a = 18

length of minor axis = 2b = 2√31

Part 3: Write The Equation Of And Graph An Ellipse.Given The Foci And Vertices Of An Ellipse, Complete
Answer 2

The various parts of the question needs to be answered.

Center: (4,-6)

Standard equation: [tex]\dfrac{(x-4)^2}{\sqrt{31}^2}+\dfrac{(y+6)^2}{9^2}=1[/tex]

Length of major axis: 18 units

Length of minor axis: [tex]2\sqrt{31}\ \text{units}[/tex]

Graph is attached.

Ellipse

The Vertices are of the form

[tex](h,k\pm a)=(4,-6\pm 9)[/tex]

This means that the ellipse is symmetrical to the y axis.

The center of an ellipse is given by [tex](h,k)[/tex]

So, here the center is [tex](4,-6)[/tex]

[tex]a=\pm 9[/tex]

The foci are of the form

[tex](h,k\pm c)=(4,-6\pm 5\sqrt{2})[/tex]

So, [tex]c=\pm 5\sqrt{2}[/tex]

[tex]b=\sqrt{a^2-c^2}\\\Rightarrow b=\sqrt{9^2-(5\sqrt{2})^2}\\\Rightarrow b=\pm\sqrt{31}[/tex]

The standard equation for major axis being parallel to y axis is given by

[tex]\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1\\\Rightarrow \dfrac{(x-4)^2}{\sqrt{31}^2}+\dfrac{(y+6)^2}{9^2}=1[/tex]

Length of major axis is

[tex]2a=2\times 9\\ =18\ \text{units}[/tex]

Length of minor axis is

[tex]2b=2\sqrt{31}\ \text{units}[/tex]

Learn more about ellipse:

https://brainly.com/question/12205108

Part 3: Write The Equation Of And Graph An Ellipse.Given The Foci And Vertices Of An Ellipse, Complete

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Step-by-step explanation:

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1. Trapezoid.

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Step-by-step explanation:

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Step-by-step explanation:

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44

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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The profit for a company this year was $100,000. Each year the profit increases at a rate of 1.2% per
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a. f(t) = 100,000 0.012^t
b. f(t) = 100,000(1 + 1.2)^t
c. f(t) = 100,000(1 + 0.012)^t
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Answers

Given:

Initial profit = $100,000

Increasing rate of profit = 1.2%

To find:

The function that shows the company's profit as a function of time in years.

Solution:

The exponential growth model is defined as:

[tex]f(t)=a(1+r)^t[/tex]

Where, a is the initial value, b is the growth rate and t is the time period.

Putting a=100,000 and r=0.012, we get

[tex]f(t)=100,000(1+0.012)^t[/tex]

Where, t is the number of years.

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There are 20 girls
and 12 boys in a class.
What would be the probability of
randomly selecting a girl? (Decimals please)

Answers

Answer:

0.625

Step-by-step explanation:

20+12=32 . ( amount of total pupils)

20/32=0.625  

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Answer:

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Step-by-step explanation:

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Answer:

10 5/8

Step-by-step explanation:

17 × 5/8  = 85/8

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Convert improper fraction to mixed number

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Answers

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Step-by-step explanation:

18b - 32

2 (9b - 16)

Answer:

2(9b-16)

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Answer:

[tex]P(32) = \frac{1}{90}[/tex]

[tex]P(Odd) = \frac{1}{2}[/tex]

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Step-by-step explanation:

Given

Sample Space = 10 to 99

First, we calculate the sample size (n)

[tex]n = 99 - 10 + 1[/tex]

[tex]n = 90[/tex]

Solving (a): P(32)

In 10 to 99, there is only 1 32.

So,

[tex]P(32) = \frac{n(32)}{n}[/tex]

[tex]P(32) = \frac{1}{90}[/tex]

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[tex]n(Odd) = 45[/tex]

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[tex]P(Odd) = \frac{n(Odd)}{n}[/tex]

[tex]P(Odd) = \frac{45}{90}[/tex]

[tex]P(Odd) = \frac{1}{2}[/tex]

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There are 18 multiples of 5 between 10 and 99

i.e

[tex]P(Multiples\ 5) = \frac{18}{90}[/tex]

[tex]P(Multiples\ 5) = \frac{1}{5}[/tex]

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Answers

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Answers

Answer:

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Step-by-step explanation:

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18/20 chance

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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