Answer:
1 circle
Step-by-step explanation:
Given two circles (red circles on the diagram).
There are two tangent circles to both of the given circles (blue circles on the diagram), and only one of them is passing through the point (0,5).
Let's check it.
The equations of the tangent circles are
![x^2+y^2=9\ [\text{Smaller tangent circle}]\\ \\x^2+y^2=25\ [\text{Larger tangent circle}]](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D9%5C%20%5B%5Ctext%7BSmaller%20tangent%20circle%7D%5D%5C%5C%20%5C%5Cx%5E2%2By%5E2%3D25%5C%20%5B%5Ctext%7BLarger%20tangent%20circle%7D%5D)
Check whether point (0,5) lies on the smaller circle:

No
Check whether point (0,5) lies on the larger circle:

Yes
<u>Answer: </u>1 circle
Answer:
1. cos(2π + θ) = cosθ
2. tan(2π + θ) = tanθ, cot(2π + θ) = cotθ.
explanation:
cot(2π + θ) = cotθ. The Trigonometric ratios of angle 2π - θ: Thinking of θ as an acute angle, 2π -θ ends in the 4th Quadrant where only cosine is positive. 2π -θ means going 1 full turn from A to A and then returning to the M' that means as much as θ.
One and one tenth is the answer
Distance of each track are:
D₁ = 428.5 yd
D₂ = 436.35 yd
D₃ = 444.20 yd
D₄ = 452.05 yd
D₅ = 459.91 yd
D₆ = 467.76 yd
D₇ = 475.61 yd
D₈ = 483.47 yd
<u>Explanation:</u>
Given:
Track is divided into 8 lanes.
The length around each track is the two lengths of the rectangle plus the two lengths of the semi-circle with varying diameters.
Thus,

Starting from the innermost edge with a diameter of 60yd.
Each lane is 10/8 = 1.25yd
So, the diameter increases by 2(1.25) = 2.5 yd each lane going outward.
So, the distances are:
D₁ = 240 + π (60) → 428.5yd
D₂ = 240 + π(60 + 2.5) → 436.35 yd
D₃ = 240 + π(60 + 5) → 444.20 yd
D₄ = 240 + π(60 + 7.5) → 452.05 yd
D₅ = 240 + π(60 + 10) → 459.91 yd
D₆ = 24 + π(60 + 12.5) → 467.76 yd
D₇ = 240 + π(60 + 15) → 475.61 yd
D₈ = 240 + π(60 + 17.5) → 483.47 yd
Answer: ( 0, 2 )
Step-by-step explanation:
You have to put coordinates of each graph on these equations y = -x + 2 and y = (1/2)x + 2. If putting coordinates satisfies both equations, then that coordinate will be the solution.
For example, let's put (0, 2) to equations.
y = -x + 2
2 = -0 + 2
2 = 2, true
y = (1/2)x + 2
2 = (1/2) × 0 + 2
2 = 2, true
So, ( 0, 2 ) is the solution.