Answer:
FJH is an obtuse angle, so it is automatically larger than 90. Answer is D.
In the given question, there are numerous information's already provided. It is important to note then down first. With the help of those given information's the required answer can be easily reached.
Percentage of students that weighed 140 pounds = 75 percent
Then
Percentage of students that weighed more than 140 pounds = (100 - 75) percent
= 25 percent
Total number of students that were weighed = 40 students
Total number of students that weighed more than 140 pounds = (25/100) * 40
= (40/4) students
= 10 students
So the number of students that weighed more than 140 pounds is 10. So the correct option in regards to the given question is option "1".
<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>
The formula for the cost of the total sales is: 8a+5c where a is the number of adult tickets and c is the number of child tickers.
If he keeps 15% the formula is: 0.15(<span>8a+5c) </span>
It would be A. 23/30 if it was not simplified but if you need it to be simplified it would be C. 1 3/20
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