Answer:
C. T is not one-to-one because the standard matrix A has a free variable.
Step-by-step explanation:
Given
Required
Determine if it is linear or onto
Represent the above as a matrix.
![T(x_1,x_2,x_3) = \left[\begin{array}{ccc}1&-5&4\\0&1&-6\\0&0&0\end{array}\right] \left[\begin{array}{c}x_1&x_2&x_3\end{array}\right]](https://tex.z-dn.net/?f=T%28x_1%2Cx_2%2Cx_3%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-5%264%5C%5C0%261%26-6%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5Cend%7Barray%7D%5Cright%5D)
From the above matrix, we observe that the matrix does not have a pivot in every column.
This means that the column are not linearly independent, & it has a free variable and as such T is not one-on-one
<span>If logc81=2 then c=9
</span>
Answer:
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Step-by-step explanation:
Given the data in the question;
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?
dA/dt = rate in - rate out
first we determine the rate in and rate out;
rate in = 3pound/gallon × 5gallons/min = 15 pound/min
rate out = A pounds/1000gallons × 5gallons/min = 5Ag/1000pounds/min
= 0.005A pounds/min
so we substitute
dA/dt = rate in - rate out
dA/dt = 15 - 0.005A
Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Answer:
Step-by-step explanation:
Given Data,
Q=23000 J
c=4.184 J/g °c
Δt= 68 °c
m=?
By using this equation,
Q=mcΔt
Answer:
well if its 2*5 then its 10 because you have to count by 5's two times and you'll get 10. I hope this helps. :3
