Answer:
482 km
63.94 degrees
Step-by-step explanation:
to solve this question we will use the cosine rule. For starters, draw your diagram. From point A, up north is 500km and 060 from there, another 300. If you join the point from the road junction back to the starting point, yoou have a triangle.
Cosine rule states that
C = 
where both A and B are the given distances, 500 and 300 respectively, C is the 3rd distance we're looking for and c is the given angle, 060
solving now, we have
C = 
C = ![\sqrt{250000 + 90000 - [215000 cos(60) }]](https://tex.z-dn.net/?f=%5Csqrt%7B250000%20%2B%2090000%20-%20%5B215000%20%20%20cos%2860%29%20%20%7D%5D)
C = ![\sqrt{340000 - [215000 * 0.5 }]](https://tex.z-dn.net/?f=%5Csqrt%7B340000%20-%20%5B215000%20%2A%200.5%20%20%7D%5D)
C = ![\sqrt{340000 - [107500 }]](https://tex.z-dn.net/?f=%5Csqrt%7B340000%20-%20%5B107500%20%20%7D%5D)
C =
C = 482 km
The bearing can be gotten by using the Sine Rule.
= 
sina/500 = sin60/482
482 sina = 500 sin60
sina = 
sina = 0.8983
a = sin^-1(0.8983)
a = 63.94 degrees
It would be A, because her annual premium rate is $3.25 for every $1,000. So 3.25x130=$422.50
A) about 14 square units
You have 11 fully filled squares 2 that are mostly filled and 2 that are about half filled
So 11+2+2(1/2)= about 14
Answer:
1. I would think it would be a appropriate to use a normal model to figure this out because sometimes when you are using stuff online you can get the same number like 20 times in a row, whereas, if you use a real dice, you most likely won't roll the same number 20 times.
2. { sorry two doesn't make sense }
Differentiating an integral removes the integral.
f(x) = integral of dt/sqrt(t^3 + 2)
f'(x) = 1/sqrt(x^3 + 2)
f'(1) = 1/sqrt(1^3 + 2)
f'(1) = 1/sqrt(3) = sqrt(3)/3.
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