X=6 and Y= 6 square root 2
<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>
Hey there!
5y + 6 - 2y
COMBINE the LIKE TERMS
= (5y - 2y) + (6)
= 3y + 6
Therefore, your answer is: 3y + 6
Good luck on your assignment and enjoy your day!
~Amphitrite1040:)
Answer:
B
Step-by-step explanation:
the all posible situations are listed in the sample space
www =won all 3 games
wwl= won first 2 games and lost the last and so on...
