Answer:
this is correct on plato/edmentum
Step-by-step explanation:
1. Rewrite the system:
x+12y=68 (i)
x=8y-12 (ii)
2. Let's substitute the equation (ii) into the equation (i):
x+12y=68
(8y-12)+12y=68
3. Then, you have:
8y-12+12y=68
4. When you clear "y", you have:
20y-12=68
20y=68+12
20y=80
y=80/20
y=4
5. You already have the value of "y". Now, you must substitute this value into the equation (ii):
x=8y-12
x=8(4)-12
x=32-12
x=20
6. Therefore, the result is:
x=20
y=4
5. arc length = radius x central angle in radians
arc length = 28 x 3π/4 = 21π
answer: (21π) cm
6. You are right. Quadrant 1 features points where both the x coordinate and the y coordinate are positive. Cosine and sine are basically special coordinates.
7. We need to find the tan(2π/3). The hint is quite bad I must say because you need to find the sine and cosine of an angle to find tangent. Okay, back to the problem.
tan(2π/3) = sin(2π/3)/cos(2π/3)
We know π/3 to be a special angle on the unit circle. It has a cosine of 1/2 and a sine of . Because we know this, its partner in quadrant 2 (2π/3) will have a cosine of -1/2 and a sine of .
tan(2π/3) = ÷ -1/2= -√3
answer: -√3
8. Both angles are special angles so...
2cos(π/6) - 2tan(π/3) = 2() - 2( ÷ 1/2) = √3 - 2√3 = -√3 (ok what a coincidence)
answer: -√3