Sketch the following figure Coplanar points E, F, and G contained in plane Q
such that plane Q cannot be named plane EFG

step-by-step explanation:

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.