An ellipse is centered at the origin. find it’s equation.

vertices: (-2,0) and (2,0)
foci: (-1,0) and (1,0)

(x)^2        (y)^2

+ = 1

 4              3

Step-by-step explanation:

The standard equation for an ellipse is

(x-h)^2     (y-k)^2

+ = 1

a^2            b^2

The center is at (h,k)

The vertices are at (h±a, k)

The foci are at (h±c,k )

Where c is sqrt(a^2 - b^2)

It is centered at the origin so h,k are zero

(x)^2        (y)^2

+ = 1

a^2            b^2

The center is at (0,0)

The vertices are at (0±a, 0)

The foci are at (0±c,0 )

The vertices are (±2,0)  so a =2

The foci is 1

c = sqrt(a^2 - b^2)

1 = sqrt(2^2 - b^2)

Square each side

1 = 4-b^2

Subtract 4 from each side

1-4 = -b^2

-3 = -b^2

3= b^2

Take the square root

b=sqrt(3)

(x)^2        (y)^2

+ = 1

 4              3

for this case we must apply distributive property to the following expression:

the distributive property states that:

then, when applying the property we have to multiply the -2 for each term within the parenthesis:

30.5*10=350$

you have to multiply the number without the decimal point then add it later but if its tenn u must move the place value of the decimal point 1 to the right.