Find the center, vertices, and foci of the ellipse with equation 3x2 + 8y2 = 24

The center is at (0,0)

The vertices are at ( ( ±2 sqrt(2),0)

foci are  ( ±sqrt(5),0)

Step-by-step explanation:

3x^2 + 8y^2 = 24

Divide each side by 24

3x^2 /24 + 8y^2/24 = 24/24

x^2/8  + y^2 /3 = 1

The general equation of an ellipse is

(x-h)^2/  a^2  + (y-k)^2 / b^2  = 1

a>b   (h,k) is the center

the coordinates of the vertices are ( ±a,0)

the coordinates of the foci are  ( ±c,0), where  ^c2=a^2−b^ 2

The center is at (0,0)

a = sqrt(8) = 2sqrt(2)

The vertices are at ( ( ±2 sqrt(2),0)

c = 8 - 3 =5

foci are  ( ±sqrt(5),0)

Centre: (0,0)

Vertices: (2sqrt(2) , 0) (-2sqrt(2) , 0)

Co-vertices: (0, sqrt(3)) (0, -sqrt(3))

Foci: (sqrt(5), 0) (-sqrt(5) , 0)

Step-by-step explanation:

(x - h)²/a² + (y - k)²/b² = 1

3x²/24 + 8y²/24 = 1

x²/8 + y²/3 = 1

Centre: (0,0)

Vertices:

y² = 3

y = +/- sqrt(3)

(0, sqrt(3))

(0, -sqrt(3))

x² = 8

x = +/- sqrt(8) = +/- 2sqrt(2)

(2sqrt(2) , 0)

(-2sqrt(2) , 0)

Foci: (c , 0)

c² = a² - b²

c² = 8 - 3 = 5

c = +/- sqrt(5)

answer is 67%

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}