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3 years ago

Graphing Functions (PLEASE HELP)

Mathematics

2 answers:

dlinn [17]3 years ago

8 0

Answer:

Which question bud? I don't k know which one you asking for

kotegsom [21]3 years ago

7 0

1. multiply each one by 5 so (1,5), (2,10), (3,15) (4, 20) (5,25)

2. multiply each by two (10,20), (20,40) (30, 60), (40, 80), (50, 100)

3. If you now get it add each by 3

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Answer:

$\frac{(x)^{2}}{49}+\frac{(y)^{2}}{1}=1$

Step-by-step explanation:

In this problem we have a horizontal ellipse, because the major axis is the x-axis

The equation of a horizontal ellipse is equal to

$\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}} =1$

where

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we have

vertices at (7, 0) and (-7, 0)

co-vertices at (0, 1) and (0, -1)

The center is the origin (0.0) (The center is the midpoint of the vertices)

a=7

b=1

substitute

$\frac{(x-0)^{2}}{7^{2}}+\frac{(y-0)^{2}}{1^{2}}=1$

$\frac{(x)^{2}}{49}+\frac{(y)^{2}}{1}=1$

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Step-by-step explanation:

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Step-by-step explanation:

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