47.5 6s in 285
I hope this helps =]
Answer:
43.17 ft/s
Step-by-step explanation:
Let the height of the wall is y.
distance of wall from light is 20 feet.
y = 20 tanθ .... (1)
time period, T = 3 second
![\frac{d\theta}{dt}=\frac{2\pi}{3}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%7D%7Bdt%7D%3D%5Cfrac%7B2%5Cpi%7D%7B3%7D)
Differentiate equation (1) with respect to t.
![\frac{dy}{dt}=20sec^{2}\theta\times \frac{d\theta}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D20sec%5E%7B2%7D%5Ctheta%5Ctimes%20%5Cfrac%7Bd%5Ctheta%7D%7Bdt%7D)
![\frac{dy}{dt}=20sec^{2}10\times \frac{2\pi}{3}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D20sec%5E%7B2%7D10%5Ctimes%20%5Cfrac%7B2%5Cpi%7D%7B3%7D)
dy/dt = 20 x 2.16
dy/dt = 43.17 ft/s
Answer:
56
Step-by-step explanation:
1:8= 7:?
1*?=7
/1 /1
1*7=7
8*7=56
1:8 = 7:56
Answer:
![\displaystyle about\:8\:units](https://tex.z-dn.net/?f=%5Cdisplaystyle%20about%5C%3A8%5C%3Aunits)
Step-by-step explanation:
Use the Distance Formula:
![\displaystyle \sqrt{[-x_1 + x_2]^2 + [-y_1 + y_2]^2} = D \\ \\ \sqrt{[6 - 1]^2 + [-7 + 1]^2} = \sqrt{5^2 + [-6]^2} = \sqrt{25 + 36} = \sqrt{61} ≈ 7,810249676 ≈ 8](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%7B%5B-x_1%20%2B%20x_2%5D%5E2%20%2B%20%5B-y_1%20%2B%20y_2%5D%5E2%7D%20%3D%20D%20%5C%5C%20%5C%5C%20%5Csqrt%7B%5B6%20-%201%5D%5E2%20%2B%20%5B-7%20%2B%201%5D%5E2%7D%20%3D%20%5Csqrt%7B5%5E2%20%2B%20%5B-6%5D%5E2%7D%20%3D%20%5Csqrt%7B25%20%2B%2036%7D%20%3D%20%5Csqrt%7B61%7D%20%E2%89%88%207%2C810249676%20%E2%89%88%208)
Since we are talking about distance, we ONLY want the NON-NEGATIVE root.
I am joyous to assist you anytime.