Consider the point. (4, 5, 6) what is the projection of the point on the xy-plane? (x, y, z) = what is the projection of the point on the yz-plane? (x, y, z) = what is the projection of the point on the xz-plane? (x, y, z) = draw a rectangular box with the origin and (4, 5, 6) as opposite vertices and with its faces parallel to the coordinate planes. label all vertices of the box.

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.

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step-by-step explanation:

tduressqwzyx86diyvkghvtufvgycyvgtytc4

using pemdas, (order of operations) you'll see the answer is 24.

there is no picture in order for this question to be answered