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)
First lets distribute the 5x
5x^2 + 30x = -50
Lets divide every term by 5
x^2 + 6x = -10
To complete the square we have to half the b value, which in this case is 6. Then square it.
Half of 6 is 3, 3 squared is 9
Add that to both sides of the equation
x^2 + 6x + 9 = -1
Find the binomial squared
(x+3)^2 (If you're wondering how i got that please comment)
(x+3)^2 = -1
Take the square root of the equation of both sides
(x+3) = +/- i
x = -3 +/- i
x = -3 - i
and
x = -3 + i
<span>The scales are different on the x-axis and the y-axis.
</span>Therefore, the correct answer choice is:
A. The x-axis scale shows the data is more clustered than it actually is.
Answer:
Mizuki here to answer! B is the correct answer!
Step-by-step explanation:
1 - 0.4 = 0.6
2.1 x 0.6 = 1.26
Answer:
The first one is ur answer
Step-by-step explanation: