Answer:
6.31 mi
Step-by-step explanation:
The diagram below explains the solution better.
From the diagram,
C = starting point of the race.
A = end of the first part of the race.
B = end of the race.
Using Cosine rule, we can find the straight-line distance between the starting point and the end of the race.
Cosine rule states that:
![a^2 = b^2 + c^2 - 2bc[cos(A)]](https://tex.z-dn.net/?f=a%5E2%20%3D%20b%5E2%20%2B%20c%5E2%20-%202bc%5Bcos%28A%29%5D)
where A = angle A = <A
Given that
b = 5.2 miles
c = 2.0 miles
<A = 115° (from the diagram)
Hence,
![a^2 = 5.2^2 + 2.0^2 - 2*5.2*2.0[cos(115)]\\\\a^2 = 27.04 + 4 - 20.8[cos(115)]\\\\a^2 = 31.04 + 8.79\\\\a^2 = 39.83\\\\a = \sqrt{39.83}\\ \\a = 6.31 mi](https://tex.z-dn.net/?f=a%5E2%20%3D%205.2%5E2%20%2B%202.0%5E2%20-%202%2A5.2%2A2.0%5Bcos%28115%29%5D%5C%5C%5C%5Ca%5E2%20%3D%2027.04%20%2B%204%20-%2020.8%5Bcos%28115%29%5D%5C%5C%5C%5Ca%5E2%20%3D%2031.04%20%2B%208.79%5C%5C%5C%5Ca%5E2%20%3D%2039.83%5C%5C%5C%5Ca%20%3D%20%5Csqrt%7B39.83%7D%5C%5C%20%5C%5Ca%20%3D%206.31%20mi)
The straight-line distance between the starting point and the end of the race is 6.31 mi
For combining like terms: it is 7z
Where are the answers?
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that acceleration of an object is

is the solution to the differential equation
Since v(0) =7
we get ln 7 = C
Hence 
since velocity is rate of change of distance s we have
![v=\frac{ds}{dt} =7e^{-2t}\\s= [tex]s(t) =\frac{-7}{2} (e^{-2t})+C)[](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bds%7D%7Bdt%7D%20%3D7e%5E%7B-2t%7D%5C%5Cs%3D%20%5Btex%5Ds%28t%29%20%3D%5Cfrac%7B-7%7D%7B2%7D%20%28e%5E%7B-2t%7D%29%2BC%29%5B)
substitute t=0 and s=0

So solution for distance is

Answer: A and B only
Step-by-step explanation:
In function, for one value of x there must be only one value of y.