Answer:
The required position of the particle at time t is: 
Step-by-step explanation:
Consider the provided matrix.



The general solution of the equation 

Substitute the respective values we get:


Substitute initial condition 

Reduce matrix to reduced row echelon form.

Therefore, 
Thus, the general solution of the equation 


The required position of the particle at time t is: 
Answer:
Step-by-step explanation:
12.9 + 10.2x = 35.1
10.2x = 35.1 - 12.9
10.2x = 22.2
x = 22.2/10.2
x= 222/102 = 111/51

Answer: C) Lindsey spent a total of $47 on her mother for her birthday. She bought a $35 box of chocolates and some balloons for $4 each. What is x, the number of balloons that Lindsey bought for her mother's birthday?
Step-by-step explanation:
Answer:
Step-by-step explanation:both are 1/6 of a chance
Answer:
7/12 or in decimal form it is .583
Step-by-step explanation:
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