I'll do the first one to get you started
The equation y = x^2+16x+64 is the same as y = 1x^2+16x+64
Compare that to y = ax^2+bx+c and we see that
a = 1
b = 16
c = 64
Use the values of 'a' and b to get the value of h as shown below
h = -b/(2a)
h = -16/(2*1)
h = -8
This is the x coordinate of the vertex.
Plug this x value into the original equation to find the corresponding y value of the vertex.
y = x^2+16x+64
y = (-8)^2 + 16(-8) + 64
y = 0
Since the y coordinate of the vertex is 0, this means k = 0.
The vertex is (h,k) = (-8, 0)
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So we found that a = 1, h = -8 and k = 0
Therefore,
f(x) = a(x-h)^2 + k
f(x) = 1(x-(-8))^2 + 0
f(x) = (x+8)^2
is the vertex form
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<h3>Final answer to problem 1 is f(x) = (x+8)^2 </h3>
Answer: There is probability that he will be asked a question in class during Week 7 is 64%.
Step-by-step explanation:
Since we have given that
Probability that question in class being asked during Week 1 = 1%
Probability that question in class being asked during Week 2 = 2%
Probability that question in class being asked during Week 3 = 4%
and so on.
So, we need to find the probability that question being asked in class during week 7.
Since it forms geometric series:
1%, 2%, 4%, ........
So, we need to find the 7 th term:

Hence, there is probability that he will be asked a question in class during Week 7 is 64%.
Im wondering the same thing i have this question to
Answer: it is 2 1/3
Step-by-step explanation:
Apply the rule: 
![3[2 ln(x-1) - lnx] + ln(x+1)=3[ln(x-1)^{2} - lnx ] + ln(x+1)](https://tex.z-dn.net/?f=3%5B2%20ln%28x-1%29%20-%20lnx%5D%20%2B%20ln%28x%2B1%29%3D3%5Bln%28x-1%29%5E%7B2%7D%20-%20lnx%20%5D%20%2B%20ln%28x%2B1%29)
Apply the rule : 
![3[2 ln(x-1) - lnx] + ln(x+1)=3ln\frac{(x-1)^{2} }{x} + ln(x+1)](https://tex.z-dn.net/?f=3%5B2%20ln%28x-1%29%20-%20lnx%5D%20%2B%20ln%28x%2B1%29%3D3ln%5Cfrac%7B%28x-1%29%5E%7B2%7D%20%7D%7Bx%7D%20%2B%20ln%28x%2B1%29)
Apply the rule: 
![3[ln (x-1)^{2} -ln x]+ln (x+1)= ln \frac{(x-1)^{6} }{x^{3} } +log(x+1)](https://tex.z-dn.net/?f=3%5Bln%20%28x-1%29%5E%7B2%7D%20-ln%20x%5D%2Bln%20%28x%2B1%29%3D%20ln%20%5Cfrac%7B%28x-1%29%5E%7B6%7D%20%7D%7Bx%5E%7B3%7D%20%7D%20%2Blog%28x%2B1%29)
Finally, apply the rule: log a + log b = log ab
![3[ln(x-1)^{2} -ln x]+log(x+1)=ln\frac{(x-1)^{6}(x+1) }{x^{3} }](https://tex.z-dn.net/?f=3%5Bln%28x-1%29%5E%7B2%7D%20-ln%20x%5D%2Blog%28x%2B1%29%3Dln%5Cfrac%7B%28x-1%29%5E%7B6%7D%28x%2B1%29%20%7D%7Bx%5E%7B3%7D%20%7D)