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1. find the standard equation of an ellipse with its foci at (2, 0) and (8,0) and a major axis of length 12.
2. find the standard equation of an ellipse with its foci at (1, 2) and (5, 2) and a major axis of length 6.
3, find the standard equation of a hyperbola with vertices (-2,0) and (2, 0), and foci (-6, 0) and (6, 0).

Step-by-step explanation:

1. Equation of an ellipse is:

(x - h)² / a² + (y - k)² / b² = 1

where (h, k) is the center and a and b are the length of half the minor/major axes.

The center is the midpoint of the foci:

(h, k) = (½ (2+8), ½(0+0))

(h, k) = (5, 0)

The foci have the same y-coordinate, so the horizontal axis is the major axis:

a = 12/2

a = 6

The distance from the foci to the center is c:

c = 8-5

c = 3

b can be found using the formula:

c² = a² - b²

3² = 6² - b²

b² = 36 - 9

b² = 27

So the equation is:

(x - 5)² / 36 + (y - 0)² / 27 = 1

2. Same steps as #1.  First find the center:

(h, k) = (½ (1+5), ½ (2+2))

(h, k) = (3, 2)

The foci have the same y-coordinate, so the horizontal axis is the major axis:

a = 6/2

a = 3

The distance from the foci to the center is c:

c = 5-3

c = 2

b can be found using the formula:

c² = a² - b²

2² = 3² - b²

b² = 9 - 4

b² = 5

So the equation is:

(x - 3)² / 9 + (y - 2)² / 5 = 1

3. The vertices have the same y coordinate, so this is a horizontal hyperbola:

(x - h)² / a² - (y - k)² / b² = 1

The center (h, k) is the midpoint of the vertices:

(h, k) = (½ (-2+2), ½ (0+0))

(h, k) = (0, 0)

The distance from the center to the vertices is a:

a = 2-0

a = 2

The distance from the center to the foci is c:

c = 6-0

c = 6

b can be found using the formula:

c² = a² + b²

6² = 2² + b²

b² = 36 - 4

b² = 32

So the equation is:

(x - 0)² / 4 - (y - 0)² / 32 = 1

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.