The given pair of functions is
![u_{1}(t) =e^{2t}, \, u_{2}(t)=e^{- \frac{3t}{2}}](https://tex.z-dn.net/?f=u_%7B1%7D%28t%29%20%3De%5E%7B2t%7D%2C%20%5C%2C%20u_%7B2%7D%28t%29%3De%5E%7B-%20%5Cfrac%7B3t%7D%7B2%7D%7D%20)
The derivatives are
![u_{1}{'}=2e^{2t} \\ u_{2}^{'} = - \frac{3}{2} e^{- \frac{3t}{2} }](https://tex.z-dn.net/?f=u_%7B1%7D%7B%27%7D%3D2e%5E%7B2t%7D%20%5C%5C%20u_%7B2%7D%5E%7B%27%7D%20%3D%20-%20%5Cfrac%7B3%7D%7B2%7D%20e%5E%7B-%20%5Cfrac%7B3t%7D%7B2%7D%20%7D)
The Wronskian is
![W=\begin{pmatrix} e^{2t} & e^{- \frac{3t}{2}}\\2e^{2t} & - \frac{3}{2}e^{- \frac{3t}{2}} \end{pmatrix}=- \frac{3}{2}e^{t/2}-2e^{t/2}=- \frac{7}{2}e^{t/2}](https://tex.z-dn.net/?f=W%3D%5Cbegin%7Bpmatrix%7D%20e%5E%7B2t%7D%20%26%20e%5E%7B-%20%5Cfrac%7B3t%7D%7B2%7D%7D%5C%5C2e%5E%7B2t%7D%20%26%20-%20%5Cfrac%7B3%7D%7B2%7De%5E%7B-%20%5Cfrac%7B3t%7D%7B2%7D%7D%20%20%20%20%5Cend%7Bpmatrix%7D%3D-%20%5Cfrac%7B3%7D%7B2%7De%5E%7Bt%2F2%7D-2e%5E%7Bt%2F2%7D%3D-%20%5Cfrac%7B7%7D%7B2%7De%5E%7Bt%2F2%7D%20%20)
Answer:
Answer: C. 4x^2/3 is ur answer
Answer:
-1/3
Step-by-step explanation:
Formula for slope is y²- y¹/x²-x¹. Plug in the numbers and you get 1-5/21-9. Do the math and you get -4/12 or -2/6 or -1/3
Answer:
(3,1) is the midpoint
Step-by-step explanation:
To find the x coordinate of the midpoint, average the x coordinates of the endpoints
(7+-1)/2 = 6/2 =3
To find the y coordinate of the midpoint, average the y coordinates of the endpoints
(10+-8)/2 = 2/2 = 1
(3,1) is the midpoint
7 is a coefficient 4 is a constant