The sum of the angles in a triangle is 180, so the equation will be :
2k+k+45= 180
add similar variables
3k+45=180
3k=135 (divide each side by 3)
k= 45
So 2k equals= 90
k=25
This is right angle triangle for your info!
Given in the problem is the diameter of the Ferris Wheel.
Thus, we can compute for the Ferris Wheel Circumference. This is the circular distance a single capsule attached to the wheel needs to do a full circle to.
Using 2 Step, we find the rate of how fast the capsule needs to be moving to complete 1 full cycle in 30 minutes.
1. Formula for computing the circumference
C = 2 x π x R
where R = Diameter divided by 2
C = 2π(120/2 )
C = 120π
2. Compute the rate or speed of the capsule / coach.
Rate or Speed = Distance to cover / Time it takes to cover
R/S = 120π/30 = 4π m/min or 12.57737 meters / min
Answer:
![10](https://tex.z-dn.net/?f=10)
Step-by-step explanation:
![\sqrt{100} \\ \sqrt{10 \times 10} \\ = 10](https://tex.z-dn.net/?f=%20%5Csqrt%7B100%7D%20%20%5C%5C%20%20%5Csqrt%7B10%20%5Ctimes%2010%7D%20%20%5C%5C%20%20%3D%2010)
hope this helps
brainliest appreciated
good luck! have a nice day!
Grayson's mistake was that he multiplied 4 and 3 and then used the exponent he had to square 3 and then multiply it by 4.
Emily's mistake was that she added 2 to 36 instead of multiplying it by -2
Pat's mistake was that he forget to make y into -2 instead of 2
The right way to do this is 4(3^2)+2(-2)
(3^2)=9 9×4=36 2(-2)=-4 -4+9=5
Answer:
34.6 units
Step-by-step explanation:
The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.
The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).
Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:
Distance between point A(-6, 2) and point B(2, 6):
![AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}](https://tex.z-dn.net/?f=%20AB%20%3D%20%5Csqrt%7B%28x_2%20-%20x_1%29%5E2%20%2B%20%28y_2%20-%20y_1%29%5E2%7D%20)
Let,
![A(-6, 2)) = (x_1, y_1)](https://tex.z-dn.net/?f=%20A%28-6%2C%202%29%29%20%3D%20%28x_1%2C%20y_1%29%20)
![B(2, 6) = (x_2, y_2)](https://tex.z-dn.net/?f=%20B%282%2C%206%29%20%3D%20%28x_2%2C%20y_2%29%20)
![AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}](https://tex.z-dn.net/?f=%20AB%20%3D%20%5Csqrt%7B%282%20-%20%28-6%29%29%5E2%20%2B%20%286%20-%202%29%5E2%7D%20)
![AB = \sqrt{(8)^2 + (4)^2}](https://tex.z-dn.net/?f=%20AB%20%3D%20%5Csqrt%7B%288%29%5E2%20%2B%20%284%29%5E2%7D%20)
![AB = \sqrt{64 + 16} = \sqrt{80}](https://tex.z-dn.net/?f=%20AB%20%3D%20%5Csqrt%7B64%20%2B%2016%7D%20%3D%20%5Csqrt%7B80%7D%20)
(nearest tenth)
Distance between B(2, 6) and C(7, 1):
![BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}](https://tex.z-dn.net/?f=%20BC%20%3D%20%5Csqrt%7B%28x_2%20-%20x_1%29%5E2%20%2B%20%28y_2%20-%20y_1%29%5E2%7D%20)
Let,
![B(2, 6) = (x_1, y_1)](https://tex.z-dn.net/?f=%20B%282%2C%206%29%20%3D%20%28x_1%2C%20y_1%29%20)
![C(7, 1) = (x_2, y_2)](https://tex.z-dn.net/?f=%20C%287%2C%201%29%20%3D%20%28x_2%2C%20y_2%29%20)
![BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}](https://tex.z-dn.net/?f=%20BC%20%3D%20%5Csqrt%7B%287%20-%202%29%5E2%20%2B%20%281%20-%206%29%5E2%7D%20)
![BC = \sqrt{(5)^2 + (-5)^2}](https://tex.z-dn.net/?f=%20BC%20%3D%20%5Csqrt%7B%285%29%5E2%20%2B%20%28-5%29%5E2%7D%20)
![BC = \sqrt{25 + 25} = \sqrt{50}](https://tex.z-dn.net/?f=%20BC%20%3D%20%5Csqrt%7B25%20%2B%2025%7D%20%3D%20%5Csqrt%7B50%7D%20)
(nearest tenth)
Distance between C(7, 1) and D(3, -5):
![CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}](https://tex.z-dn.net/?f=%20CD%20%3D%20%5Csqrt%7B%28x_2%20-%20x_1%29%5E2%20%2B%20%28y_2%20-%20y_1%29%5E2%7D%20)
Let,
![C(7, 1) = (x_1, y_1)](https://tex.z-dn.net/?f=%20C%287%2C%201%29%20%3D%20%28x_1%2C%20y_1%29%20)
![D(3, -5) = (x_2, y_2)](https://tex.z-dn.net/?f=%20D%283%2C%20-5%29%20%3D%20%28x_2%2C%20y_2%29%20)
![CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}](https://tex.z-dn.net/?f=%20CD%20%3D%20%5Csqrt%7B%283%20-%207%29%5E2%20%2B%20%28-5%20-%201%29%5E2%7D%20)
![CD = \sqrt{(-4)^2 + (-6)^2}](https://tex.z-dn.net/?f=%20CD%20%3D%20%5Csqrt%7B%28-4%29%5E2%20%2B%20%28-6%29%5E2%7D%20)
![CD = \sqrt{16 + 36} = \sqrt{52}](https://tex.z-dn.net/?f=%20CD%20%3D%20%5Csqrt%7B16%20%2B%2036%7D%20%3D%20%5Csqrt%7B52%7D%20)
(nearest tenth)
Distance between D(3, -5) and A(-6, 2):
![DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}](https://tex.z-dn.net/?f=%20DA%20%3D%20%5Csqrt%7B%28x_2%20-%20x_1%29%5E2%20%2B%20%28y_2%20-%20y_1%29%5E2%7D%20)
Let,
![D(3, -5) = (x_1, y_1)](https://tex.z-dn.net/?f=%20D%283%2C%20-5%29%20%3D%20%28x_1%2C%20y_1%29%20)
![A(-6, 2) = (x_2, y_2)](https://tex.z-dn.net/?f=%20A%28-6%2C%202%29%20%3D%20%28x_2%2C%20y_2%29%20)
![DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}](https://tex.z-dn.net/?f=%20DA%20%3D%20%5Csqrt%7B%28-6%20-%203%29%5E2%20%2B%20%282%20-%20%28-5%29%29%5E2%7D%20)
![DA = \sqrt{(-9)^2 + (7)^2}](https://tex.z-dn.net/?f=%20DA%20%3D%20%5Csqrt%7B%28-9%29%5E2%20%2B%20%287%29%5E2%7D%20)
![DA = \sqrt{81 + 49} = \sqrt{130}](https://tex.z-dn.net/?f=%20DA%20%3D%20%5Csqrt%7B81%20%2B%2049%7D%20%3D%20%5Csqrt%7B130%7D%20)
(nearest tenth)
Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units