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3 years ago

Quick please help!!!!

Mathematics

1 answer:

Westkost [7]3 years ago

4 0

From the equation, we have

$h=-20, k =21 , a^2=20^2, b^2=21^2$

To find the vertex and focus, we have to use the following formula

$Vertices =(h, k+-a) \\ Focii = (h,k+-c)$

And

$c = \sqrt{a^2 + b^2} = \sqrt{400+441} = 29$

So, we have

$Vertices=(-21,20+-20) \\ Focii=(-21, 20+-29)$

$Vertices =(-21,0),(-21,40) \\ focii=(-21,-9),(-21,49)$

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Answer:

(x+1)^2+(y+1)^2=13

Step-by-step explanation:

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center: (-1, -1)

radius: sqrt(6^2+4^2)/2=sqrt(52)/2=2sqrt(13)/2=sqrt(13)

Substitute those values in to get

(x+1)^2+(y+1)^2=13

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<h2>Hello!</h2>

The answer is:

The polynomial that can be simplified to a difference of squares is the second polynomial:

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To solve this problem, we need to look for which of the given quadratic terms given for the different polynomials can be a result of squaring (elevating by two).

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Discarding, we have:

The quadratic terms of the given polynomials are:

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So, the first and the fourth polynomial are discarded and cannot be simplified to a difference of squares at least using whole numbers.

Therefore, we need to work with the second and the third polynomial.

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$16a^{2}-4a+4a-1=16a^{2}=(4a)^{2}-(1)^{2}$

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