Answer:
66 2/3 as a mixed number or 200/3 as an improper fraction.
Step-by-step explanation:
That would be 66 2/3.
0.6 recurring is 2/3.
<span>Assuming that the particle is the 3rd
particle, we know that it’s location must be beyond q2; it cannot be between q1
and q2 since both fields point the similar way in the between region (due to
attraction). Choosing an arbitrary value of 1 for L, we get </span>
<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>
Rearranging to calculate for d:
<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0
d = 2.72075922005613
d = 0.612574113277207 </span>
<span>
We pick the value that is > q2 hence,</span>
d = 2.72075922005613*L
<span>d = 2.72*L</span>
Answer:
Step-by-step explanation:
Let x represent the number of years it will take the two colleges to have the same enrollment.
In 2000, there were 12900 students at college A, with a projected enrollment increase of 900 students per year. This means that the expected number of students at college A in x years time is
12900 + 900x
In the same year, there were 25,000 students at college B, with a projected enrollment decline of 700 students per year. This means that the expected number of students at college B in x years time is
25000 - 700x
For both colleges to have the same enrollment,
12900 + 900x = 25000 - 700x
900x + 700x = 25000 - 12900
1600x = 12100
x = 12100/1600
x = 7.56
Approximately 8 years
The year would be 2000 + 8 = 2008
