Answer:
Distance to the xy-plane = |z|
Distance to the yz-plane = |x|
Distance to the xz-plane = |y|
Step-by-step explanation:
The distance from P(x,y,z) to the xy-plane is by definition the magnitude of the vector that goes from the perpendicular projection of P over the xy-plane to the point P, which is exactly the magnitude of the vector (0,0,z) = |z| the absolute value of z
Similarly, the distance from P to the yz-plane is |x| and the distance from P to the xz-plane is |y|
Distance to the xy-plane = |z|
Distance to the yz-plane = |x|
Distance to the xz-plane = |y|
The answer is 4 units. The distance between the two points of B and C is equals 6-4=2, because the y is the same. According to the question, the BC is equals 0.5*B'C'. So the length of B'C' is 2/0.5=4 units.
46 because if the age of the son was 8 then the father was 24 and if we add 8 (because that’s how many years passed) we get 16 and 32 which fit the criteria so if we take off 1 year on each (because only 7 years pass instead of 8) we can add them together and get 46
The answer would be 6.3, Three squared is nine and the missing side needs to equal 49. 6.3 squares and 3 squared will equal 49.
Answer: C
Step-by-step explanation:
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