Lets first simplify the dimensions of the rectangle
length = 7
breadth =
=
= 0.6
The area of rectangle = length * breadth
so, area of rectangle = 7 * 0.6
Area of rectangle = 4.2
Hope it helped you.. :D
Answer: k=9
reasoning: just plug in 1 for x and 1 for y
Notice the graph, the domain is just the horizontal area "used up" over the x-axis, so, the graph goees from
![\bf -\cfrac{5x}{2}\quad to\quad \cfrac{5x}{2}\implies domain\implies \left[-\cfrac{5x}{2}\ ,\ \cfrac{5x}{2}\right]](https://tex.z-dn.net/?f=%5Cbf%20-%5Ccfrac%7B5x%7D%7B2%7D%5Cquad%20to%5Cquad%20%5Ccfrac%7B5x%7D%7B2%7D%5Cimplies%20domain%5Cimplies%20%5Cleft%5B-%5Ccfrac%7B5x%7D%7B2%7D%5C%20%2C%5C%20%20%20%5Ccfrac%7B5x%7D%7B2%7D%5Cright%5D)
the range is just, the vertical area "used up" over the y-axis, so, the graph goes to 2 and down to -2, thus
![\bf range\implies [2\ ,\ -2]](https://tex.z-dn.net/?f=%5Cbf%20range%5Cimplies%20%5B2%5C%20%2C%5C%20-2%5D)
This situation is a "union." To answer it, add together the probabilities 0.42 and 0.38. Result: 0.80.
The probability of getting into either USC or VSU is 0.80.
The general solutions always have some additive/multiplicative constant, that you must fix in the particular solution.
In order to do so, you need to impose that the particular solution passes through a certain point. In your case, you have
and you want
Put everything together, and you have
Since the cosine is zero in the chosen point. So, we've fixed the value of the constant, and the particular solution is found:
