Answer:
Part A
The bearing of the point 'R' from 'S' is 225°
Part B
The bearing from R to Q is approximately 293.2°
Step-by-step explanation:
The location of the point 'Q' = 35 km due East of P
The location of the point 'S' = 15 km due West of P
The location of the 'R' = 15 km due south of 'P'
Part A
To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that 
= 
, the right triangle ΔRPS is an isosceles right triangle
∴ ∠PRS = ∠PSR = 45°
The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°
Part B
∠PRQ = arctan(35/15) ≈ 66.8°
Therefore the bearing from R to Q = 270 + 90 - 66.8 ≈ 293.2°
Answer:
no Solution
Step-by-step explanation:
-12x-12y=4\\ 3x+3y=0
12x-12y=4
add 12y to both sides
12x-12y+12y=4+12y
divid both sides by -12
\frac{-12x}{-12}=\frac{4}{-12}+\frac{12y}{-12}
simplfy
x=-\frac{1+3y}{3}
\mathrm{Substitute\:}x=-\frac{1+3y}{3}
\begin{bmatrix}3\left(-\frac{1+3y}{3}\right)+3y=0\end{bmatrix}
\begin{bmatrix}-1=0\end{bmatrix}
When you rotate the ef counter lock 180 degrees, the new line e'f' will have the same length as ef.
Answer:
Step-by-step explanation:
Given
--- interval
Required
The probability density of the volume of the cube
The volume of a cube is:
For a uniform distribution, we have:
and
implies that:
So, we have:
Solve
Recall that:
Make x the subject
So, the cumulative density is:
becomes
The CDF is:
Integrate
![F(x) = [v]\limits^{v^\frac{1}{3}}_9](https://tex.z-dn.net/?f=F%28x%29%20%3D%20%5Bv%5D%5Climits%5E%7Bv%5E%5Cfrac%7B1%7D%7B3%7D%7D_9)
Expand
The density function of the volume F(v) is:
Differentiate F(x) to give:
So:
