This is a case of direct variation.
For each hour traveled, the car traveled 50 miles.
d = 50t
Try this suggested option, answers are marked with colour.
Answer:
3×(27-7) says to subtract 7 from 27 first, then multiply with 3; but 3×27-7 says to multiply 3 by 27 first ,then subtract 7.
The second bullet is the correct option
Answer:
At 43.2°.
Step-by-step explanation:
To find the angle we need to use the following equation:
![d*sin(\theta) = m\lambda](https://tex.z-dn.net/?f=%20d%2Asin%28%5Ctheta%29%20%3D%20m%5Clambda%20)
Where:
d: is the separation of the grating
m: is the order of the maximum
λ: is the wavelength
θ: is the angle
At the first-order maximum (m=1) at 20.0 degrees we have:
![\frac{\lambda}{d} = \frac{sin(\theta)}{m} = \frac{sin(20.0)}{1} = 0.342](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Clambda%7D%7Bd%7D%20%3D%20%5Cfrac%7Bsin%28%5Ctheta%29%7D%7Bm%7D%20%3D%20%5Cfrac%7Bsin%2820.0%29%7D%7B1%7D%20%3D%200.342%20)
Now, to produce a second-order maximum (m=2) the angle must be:
![sin(\theta) = \frac{\lambda}{d}*m](https://tex.z-dn.net/?f=%20sin%28%5Ctheta%29%20%3D%20%5Cfrac%7B%5Clambda%7D%7Bd%7D%2Am%20)
![\theta = arcsin(\frac{\lambda}{d}*m) = arcsin(0.342*2) = 43.2 ^{\circ}](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20arcsin%28%5Cfrac%7B%5Clambda%7D%7Bd%7D%2Am%29%20%3D%20arcsin%280.342%2A2%29%20%3D%2043.2%20%5E%7B%5Ccirc%7D%20)
Therefore, the diffraction grating will produce a second-order maximum for the light at 43.2°.
I hope it helps you!
Step-by-step explanation:
△BAC ≅ △EDF (given)
Therefore,
BC=EF
x-4=19
x=23