I put it them on a graph, they look like a straight line to me. That makes it Linear!
Answer:
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)
For the first problem, the triangle is equilateral. This means that all three sides will have the same length. So you need to find a value for x that makes all three of those statements equivalent. I believe you could find it using a system of equations... but it’s been a while so don’t quote me on that.
For the second, ce and de are equal. Set them equal to each other and solve for x. Then plug that x value back into the side lengths.
9514 1404 393
Answer:
72
Step-by-step explanation:
h(4) = 4^3 -2·4 = 64 -8 = 56
g(4) = 3·4 +4 = 16
Then the sum is ...
(h+g)(4) = h(4) + g(4) = 56 +16
(h+g)(4) = 72
