A clock maker views a clock as a unit circle. To determine the placement of each number he needs to use the unit circle to determine the terminal point of each number on the face of the clock. What are the terminal points of each hour on the clock.

1,2,3,4,5,6,7,8,9,10,11,12 o'clock

ex) for 3 o'clock the terminal point is (1, 0)

c

step-by-step explanation:

at least i think its

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.

you could deduce the conclusion directly using the facts pertaining to the specific situation.

hope this .