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
I think it should be B because the depth of the water should be decreasing slowing at the same rate.
Answer:
a)
a1 = log(1) = 0 (2⁰ = 1)
a2 = log(2) = 1 (2¹ = 2)
a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849
a4 = log(4) = 2 (2² = 4)
a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322
a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)
a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807
a8 = log(8) = 3 (2³ = 8)
a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )
a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322
b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:
log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that
a1 = 1
a2 = 2
a3 = 3
a4 = 4
a5 = 5
a6 = 6
a7 = 7
a8 = 8
a9 = 9
a10 = 10
I hope this works for you!!
Answer:
A = 40/n
Therefore, she gave each friend 40/n ounces of jelly beans.
Step-by-step explanation:
Given;
Total amount of jelly beans To = 3 pounds = 3×16 ounces = 48 ounces
Total amount kept for family Ta = 8 ounces
Total amount given to friends = Ti
Number of friends = n
Amount given to each friend = A
Ti = To - Ta
A = Ti/n = (To-Ta)/n
A = (48 - 8)/n
A = 40/n
Therefore she gave each friend 40/n ounces of jelly beans.
The best way to solve a problem like this is guess and check.
y = 2x + 10
2 = 2(2) +10
2 = 4+10
2 = 14 NO
y = 2x + 10
4 = -3(2) +10
4 = -6 +10
4 = 4 YES . . . C is your answer
