1. consider the surface x^2 + y^2 + 9z^2 = 13 and the surface 1 = 2. (a) (2 points) classify each surface. (b) (2 points) sketch x + y^2 +9z^2 = 13. (c) (3 points) find a vector-valued function representing their intersection. (d) (2 points) find the arc length of this parametrization.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

c) yes, because of asa or aas

this is a good question! i'll do some research and come back when i have the answer!