Doubling formula is this:
![P(t)=P(2)^{\frac{t}{d}}](https://tex.z-dn.net/?f=P%28t%29%3DP%282%29%5E%7B%5Cfrac%7Bt%7D%7Bd%7D%7D)
where P=initial number of rabbits
t=time
d=time it takes to doulbe
ok, so 4 weeks is the doubling time so that is 4*7=28 days
we wawnt time=98
and oroiginal number of rabbits is 5 so
![P(98)=5(2)^{\frac{98}{28}}](https://tex.z-dn.net/?f=P%2898%29%3D5%282%29%5E%7B%5Cfrac%7B98%7D%7B28%7D%7D)
![P(98)=5(2)^{3.5}](https://tex.z-dn.net/?f=P%2898%29%3D5%282%29%5E%7B3.5%7D)
![P(98)=5(2^3)(\sqrt{2})](https://tex.z-dn.net/?f=P%2898%29%3D5%282%5E3%29%28%5Csqrt%7B2%7D%29)
![P(98)=5(8)\sqrt{2}](https://tex.z-dn.net/?f=P%2898%29%3D5%288%29%5Csqrt%7B2%7D)
![P(98)=40\sqrt{2}](https://tex.z-dn.net/?f=P%2898%29%3D40%5Csqrt%7B2%7D)
so P(98)≈56.56
we can't have .56 rabbit so round down or up
about 56 or 57 rabbits in 98 days
Answer:
I believe the first question is the third option. I hope this helps!
We will simplify the left hand side of the equation to look like the right
![\displaystyle \frac{1-sin(t)}{cos^{2}(t)} + \frac{1}{1 - sin(t)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1-sin%28t%29%7D%7Bcos%5E%7B2%7D%28t%29%7D%20%20%2B%20%5Cfrac%7B1%7D%7B1%20-%20sin%28t%29%7D)
taking the LCM
![\displaystyle \frac{[1-sin(t)]^{2} + cos^{2}(t)}{[cos^{2}(t)][1 - sin(t)]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5B1-sin%28t%29%5D%5E%7B2%7D%20%2B%20cos%5E%7B2%7D%28t%29%7D%7B%5Bcos%5E%7B2%7D%28t%29%5D%5B1%20-%20sin%28t%29%5D%7D)
expanding the binomial in the numerator
![\displaystyle \frac{[1 + sin^{2}(t) - 2sin(t) + cos^{2}(t)]}{[cos^{2}(t)][1 - sin(t)]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5B1%20%2B%20sin%5E%7B2%7D%28t%29%20-%202sin%28t%29%20%2B%20cos%5E%7B2%7D%28t%29%5D%7D%7B%5Bcos%5E%7B2%7D%28t%29%5D%5B1%20-%20sin%28t%29%5D%7D)
Since sin²x + cos²x = 1
![\displaystyle \frac{[2 - 2sin(t)]}{[cos^{2}(t)][1 - sin(t)]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5B2%20-%202sin%28t%29%5D%7D%7B%5Bcos%5E%7B2%7D%28t%29%5D%5B1%20-%20sin%28t%29%5D%7D)
factoring out the 2 from the numerator
![\displaystyle \frac{2[1 - sin(t)]}{[cos^{2}(t)][1 - sin(t)]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B2%5B1%20-%20sin%28t%29%5D%7D%7B%5Bcos%5E%7B2%7D%28t%29%5D%5B1%20-%20sin%28t%29%5D%7D)
1-sin(t) will cancel out
![\displaystyle \frac{2}{cos^{2}(t)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B2%7D%7Bcos%5E%7B2%7D%28t%29%7D)
since cos²(t) = 1/(sec²(t))
![2 sec^{2}(t)](https://tex.z-dn.net/?f=2%20sec%5E%7B2%7D%28t%29)
which is equal to the right hand side of our given equation.
So we verified the identity!
Answer:
D
0.4
Step-by-step explanation:
P(A)=0.6
P(B)=0.3
P(B∩A)=0.5
P(A∪B)=P(A)+P(B)-P(A∩B)
P(A or B)=P(A∪B)=0.6+0.3-0.5=0.9-0.5=0.4
The answer would be 100/101 if we follow according to the rule on which the sequence is based on!