Answer:
y=ln(x/(1-x))
Step-by-step explanation:
y=e^x/(1+e^x)
Cross multiply
y(1+e^x)=e^x
Distribute
y+ye^x=e^x
Put anything with x on with side and everything without x on opposing side:
y=e^x-ye^x
Factor right hand side
y=(1-y)e^x
Divide both sides by (1-y)
y/(1-y)=e^x
Use natural log.
ln(y/(1-y))=x
The inverse is
y=ln(x/(1-x))
What are you looking for?? Area? Perimeter?
Answer:
3
Step-by-step explanation:
lim(t→∞) [t ln(1 + 3/t) ]
If we evaluate the limit, we get:
∞ ln(1 + 3/∞)
∞ ln(1 + 0)
∞ 0
This is undetermined. To apply L'Hopital's rule, we need to rewrite this so the limit evaluates to ∞/∞ or 0/0.
lim(t→∞) [t ln(1 + 3/t) ]
lim(t→∞) [ln(1 + 3/t) / (1/t)]
This evaluates to 0/0. We can simplify a little with u substitution:
lim(u→0) [ln(1 + 3u) / u]
Applying L'Hopital's rule:
lim(u→0) [1/(1 + 3u) × 3 / 1]
lim(u→0) [3 / (1 + 3u)]
3 / (1 + 0)
3
You have x , x+7, and 7x-7. When you add those up and equal it to 180 you can get your x. X equals 20
So one angle is 20 degrees
The second angle is 27 degrees
And the third angle is 133 degrees.
The probability of event A and B to both occur is denoted as P(A ∩ B) = P(A) P(B|A). It is the probability that Event A occurs times the probability that Event B occurs, given that Event A has occurred.
So, to find the probability that you will be assigned a poem by Shakespeare and by Tennyson, let Event A = the event that a Shakespeare poem will be assigned to you; and let Event B = the event that the second poem that will be assigned to you will be by Tennyson.
At first, there are a total of 13 poems that would be randomly assigned in your class. There are 4 poems by Shakespeare, thus P(A) is 4/13.
After the first selection, there would be 13 poems left. Therefore, P(B|A) = 2/12
Based on the rule of multiplication,
P(A ∩ B) = P(A) P(B|A)P(A ∩ B) = 4/13 * 2/12
P(A ∩ B) = 8/156
P(A ∩ B) = 2/39
The probability that you will be assigned a poem by Shakespeare, then a poem by Tennyson is 2/39 or 5.13%.