Answer:
Step-by-step explanation:
3 ln(p + q) − 2 ln(q) − 7 ln(p)
=ln(p+q)^3- ln(q)^2-ln(p)^7
= ln(p+q)^3-{ln(q)^2+ln(p)^7}
=ln(p+q)^3-ln(q)^2*(p)^7
=ln{ (p+q)^3÷(q)^2*(p)^7}
=
Answer:
The probability that a book picked at random is either a work written by an author who writes only nonfiction or a work written by a man is 75.8%.
Step-by-step explanation:
The probability, in this case, has to be calculated using the sum rule because the events are independent.
P = P(men) + P(non-fiction) - P(men non-fiction); where P is the probability. It means the probability that the random book was written by a man plus the probability that the random book was non-fiction minus the probability that the random book was written by a man who writes non-fiction. The probability that the random book was written by a man who writes non-fiction has to be subtracted, if not you will be duplicating it.
Facts:
P(men) = 60% = 99/165
P(non-fiction) = 40% = 66/165
P(men non-fiction) = 40/165
P = P(men) + P(non-fiction) – P(men non-fiction)
P = 99/165 + 66/165 – 40/165
P = 125/165
P = 0.758 or 75.8%
Your absolute value is x+5 so next you would find the answer to 5 minus 8 which is -3 but since -3 and x are on the same side of the equation you have to add 3 to both sides. So the answer is 3
Below is the solution, I hope it helps.
<span>i) tan(70) - tan(50) = tan(60 + 10) - tan(60 - 10)
= {tan(60) + tan(10)}/{1 - tan(60)*tan(10)} - {tan(60) - tan(10)}/{1 + tan(10)*tan(60)}
ii) Taking LCM & simplifying with applying tan(60) = √3, the above simplifies to:
= 8*tan(10)/{1 - 3*tan²(10)}
iii) So tan(70) - tan(50) + tan(10) = 8*tan(10)/{1 - 3*tan²(10)} + tan(10)
= [8*tan(10) + tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)}
= [9*tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)}
= 3 [3*tan(10) - tan³(10)]/{1 - 3*tan²(10)}
= 3*tan(30) = 3*(1/√3) = √3 [Proved]
[Since tan(3A) = {3*tan(A) - tan³(A)}/{1 - 3*tan²(A)},
{3*tan(10) - tan³(10)}/{1 - 3*tan²(10)} = tan(3*10) = tan(30)]</span>
