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iragen [17]
3 years ago
6

Help me fast please

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
3 0

Answer:

\frac{x^2}{169} +\frac{y^2}{25}=1

If we compare this to the general expression for an ellipse given by:

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

We can see that the vertex is V=(0,0)

And we can find the values of a and b like this:

a=\sqrt{169}=13, b=\sqrt{25}=5

in order to find the foci we can find the value of c

c =\sqrt{169-25}=\sqrt{144}=12

The two focis are (12,0) and (-12,0)

The convertices  for this case are: (13,0) and (-13,0) on the x axis

And for the y axis (0,5) and (0,-5)

Step-by-step explanation:

For this problem we have the following equation given:

\frac{x^2}{169} +\frac{y^2}{25}=1

If we compare this to the general expression for an ellipse given by:

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

We can see that the vertex is V=(0,0)

And we can find the values of a and b like this:

a=\sqrt{169}=13, b=\sqrt{25}=5

in order to find the foci we can find the value of c

c =\sqrt{169-25}=\sqrt{144}=12

The two focis are (12,0) and (-12,0)

The convertices  for this case are: (13,0) and (-13,0) on the x axis

And for the y axis (0,5) and (0,-5)

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Table


x      f(x) =  2^x             h(x) = x^3 + x + 8

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