78 14.0.1 Find the equation for the tangent plane and the normal line at the point Po(1,2,3) on the surface 2x2 + 3y2 + 4z2 = 50. Using a coefficient of 1 for x, the equation for the tangent plane is​

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

answer: .75 km

step-by-step explanation:

the answer is a.

135 and 80

step-by-step explanation:

there is no points to select.

step-by-step explanation: