Answer:
n = 3 + i√11 , 3 - i√11
Step-by-step explanation:
This is a quadratic equation and the first step is to write it in standard form:
5n^2 - 30n + 100 = 0
Divide through by 5:
n^2 - 6n + 20 = 0
Completing the square:-
(n - 3)^2 - 9 + 20 = 0
(n - 3)^2 = -11
n - 3 = i√11 or - i√11
n = 3 + i√11 , 3 - i√11
(2 complex roots).
Answer:
(2) / ((x + 5)(x - 1))
&
(3x) / ((x - 6)(x - 1))
Step-by-step explanation:
The question is asking you to factor the denominator. In this case, you cannot simplify either of the numerators. You can factor the denominator by asking yourself the question: which two numbers multiply to equal the final number in the quadratic and add to equal the middle number in the quadratic?
1.) 2 / (x² + 4x - 5)
-----> Which two numbers multiply to -5 and add to 4? These numbers are 5 and -1. This makes the factors (x + 5) and (x - 1). The fully simplified answer: (2) / ((x + 5)(x - 1))
2.) 3x / (x² - 7x + 6)
------> Which two numbers multiply to 6 and add to -7? These numbers are -6 and -1. This makes the factors (x - 6) and (x - 1). The fully simplified answer: (3x) / ((x - 6)(x - 1))
*Please let me know if I misinterpreted the question or you need further clarification <3
Answer:
The answer is "9 and 7".
Step-by-step explanation:
Given:
![A=\left[\begin{array}{cc}7&9\\0&9\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D)
Using formula:
![|A-\lambda \cdot I|= 0\\\\](https://tex.z-dn.net/?f=%7CA-%5Clambda%20%5Ccdot%20I%7C%3D%200%5C%5C%5C%5C)
![\to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]-\lambda \left[\begin{array}{cc}1&0\\0&1\end{array}\right] |=0\\\\\\ \to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]- \left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right] |=0\\\\\\\to|\left[\begin{array}{cc}7-\lambda &9\\0&9-\lambda\end{array}\right]|=0\\\\\\\to|(7-\lambda)(9-\lambda)|=0\\\\\to (7-\lambda)(9-\lambda)=0\\\\\to 7-\lambda=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9-\lambda=0\\\\](https://tex.z-dn.net/?f=%5Cto%20%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D-%5Clambda%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%7C%3D0%5C%5C%5C%5C%5C%5C%20%5Cto%20%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D-%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%7C%3D0%5C%5C%5C%5C%5C%5C%5Cto%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7-%5Clambda%20%269%5C%5C0%269-%5Clambda%5Cend%7Barray%7D%5Cright%5D%7C%3D0%5C%5C%5C%5C%5C%5C%5Cto%7C%287-%5Clambda%29%289-%5Clambda%29%7C%3D0%5C%5C%5C%5C%5Cto%20%287-%5Clambda%29%289-%5Clambda%29%3D0%5C%5C%5C%5C%5Cto%207-%5Clambda%3D0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%209-%5Clambda%3D0%5C%5C%5C%5C)
![\to \lambda=7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda=9\\\\](https://tex.z-dn.net/?f=%5Cto%20%5Clambda%3D7%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5Clambda%3D9%5C%5C%5C%5C)
Answer:
D. 18 + (-7)
Step-by-step explanation:
18 - 7 = 11
18 + (-7) = 11
A, B, and C are wrong because...
A. -25
B. 25
C. -11
Hope this helps!! <3
Explanation
The model has the form
![y=ae^{-kt}](https://tex.z-dn.net/?f=y%3Dae%5E%7B-kt%7D)
Where a=initial amount
y= final amount
K= growth rate constant
t= time
When 140 kg of substance is left after 7 hours, the formula can be remodeled to be.
![\begin{gathered} 140=400e^{-7k} \\ e^{-7k}=\frac{140}{400} \\ e^{-7k}=\frac{7}{20} \\ \ln (e^{-7k})=\ln (\frac{7}{20}) \\ -7k=\ln (\frac{7}{20}) \\ k=\frac{\ln(\frac{7}{20})}{-7} \\ \therefore k=\frac{\ln (\frac{20}{7})}{7} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20140%3D400e%5E%7B-7k%7D%20%5C%5C%20e%5E%7B-7k%7D%3D%5Cfrac%7B140%7D%7B400%7D%20%5C%5C%20e%5E%7B-7k%7D%3D%5Cfrac%7B7%7D%7B20%7D%20%5C%5C%20%5Cln%20%28e%5E%7B-7k%7D%29%3D%5Cln%20%28%5Cfrac%7B7%7D%7B20%7D%29%20%5C%5C%20-7k%3D%5Cln%20%28%5Cfrac%7B7%7D%7B20%7D%29%20%5C%5C%20k%3D%5Cfrac%7B%5Cln%28%5Cfrac%7B7%7D%7B20%7D%29%7D%7B-7%7D%20%5C%5C%20%5Ctherefore%20k%3D%5Cfrac%7B%5Cln%20%28%5Cfrac%7B20%7D%7B7%7D%29%7D%7B7%7D%20%5Cend%7Bgathered%7D)
Therefore, the first solution is
![y=400e^{-\ln (\frac{20}{7})\frac{t}{7}}](https://tex.z-dn.net/?f=y%3D400e%5E%7B-%5Cln%20%28%5Cfrac%7B20%7D%7B7%7D%29%5Cfrac%7Bt%7D%7B7%7D%7D)
For part b we have 16 hours.
![\begin{gathered} y=400e^{-\ln (\frac{20}{7})\frac{t}{7}}=400e^{-\ln (\frac{20}{7})\frac{16}{7}} \\ y=36.302\approx36\operatorname{kg}\text{ (To the nearest whole number)} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D400e%5E%7B-%5Cln%20%28%5Cfrac%7B20%7D%7B7%7D%29%5Cfrac%7Bt%7D%7B7%7D%7D%3D400e%5E%7B-%5Cln%20%28%5Cfrac%7B20%7D%7B7%7D%29%5Cfrac%7B16%7D%7B7%7D%7D%20%5C%5C%20y%3D36.302%5Capprox36%5Coperatorname%7Bkg%7D%5Ctext%7B%20%28To%20the%20nearest%20whole%20number%29%7D%20%5Cend%7Bgathered%7D)
Thus, the answer is 36kg