Answer:
40%
Step-by-step explanation:
1.50 - 0.90 = 0.60
0.60 cents are saved.
0.4 = 40%
hope this helps :)
Please do number 19 for me.
2 answers:
You might be interested in
Answer:
To this question; we want to show that if two antipodal points on a sphere were A and B, for any random point C on the sphere, AC is perpendicular to BC?
Step-by-step explanation:
I would proceed by imagining the sphere in a three dimensional Cartesian coordinate system. For example, use a sphere of diameter 2 and let it sit at the origin. Then (0, 0, 1) and (0, 0, -1) are the vector locations of the “north and south poles” of the sphere.
Now choose the vector location of any point on the surface of the sphere. It will have a vector location - we’ll call it (x, y, z). Now the vectors from your point to the two poles are (-x, -y, 1-z) and (-x, -y, -1-z).
Now just form the dot product of those two vectors:
(-x, -y, 1-z) . (-x, -y, -1-z) = x^2 +y^2 + (1-z)*(-1-z)
Now the truth of your claim will be embodied in that dot product being zero:
x^2 + y^2 - (1+z)(1-z) = 0
x^2 + y^2 - (1-z^2) = 0
x^2 + y^2 + z^2 = 1
But that last line is just the definition of points on the surface of a sphere of radius 1, so the claim is proven.
Since R>0 and AC is perpendicular to BC, the <ACB is at right angle.
Please, find attached a simple image to show antipodal points (Two points that makes a diameter) and a point C on the sphere.
