Answer:
The equation that represents the motion of the string is given by:
.....[1] where t represents the time in second.
Given that: A = 0.6 cm (distance above its resting position) , k = 1.8(damping constant) and frequency(f) = 105 cycles per second.
Substitute the given values in [1] we get;
or
(a)
The trigonometric function that models the motion of the string is given by:

(b)
Determine the amount of time t that it takes the string to be damped so that 
Using graphing calculator for the equation
let x = t (time in sec)
Graph as shown below in the attachment:
we get:
the amount of time t that it takes the string to be damped so that
is, 0.5 sec
150 x 5= 750
20 x 5= 100 ... so they should expect 100 cellphones to be defective
Answer:
Growth
Step-by-step explanation:
Answer:
64
π
(
c
m
)
^2
Step-by-step explanation:
Answer:
See explanation
Step-by-step explanation:
Given:
![A=\left[\begin{array}{cc}-2&4\\1&3\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%264%5C%5C1%263%5Cend%7Barray%7D%5Cright%5D)
![B=\left[\begin{array}{cc}-2&1\\3&7\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%261%5C%5C3%267%5Cend%7Barray%7D%5Cright%5D)
A. Find AB:
![AB=\left[\begin{array}{cc}-2&4\\1&3\end{array}\right]\cdot \left[\begin{array}{cc}-2&1\\3&7\end{array}\right]=\left[\begin{array}{cc}-2\cdot (-2)+4\cdot 3&-2\cdot 1+4\cdot 7\\1\cdot (-2)+3\cdot 3&1\cdot 1+3\cdot 7\end{array}\right]=\left[\begin{array}{cc}16&26\\7&22\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%264%5C%5C1%263%5Cend%7Barray%7D%5Cright%5D%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%261%5C%5C3%267%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%5Ccdot%20%28-2%29%2B4%5Ccdot%203%26-2%5Ccdot%201%2B4%5Ccdot%207%5C%5C1%5Ccdot%20%28-2%29%2B3%5Ccdot%203%261%5Ccdot%201%2B3%5Ccdot%207%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D16%2626%5C%5C7%2622%5Cend%7Barray%7D%5Cright%5D)
B. Find BA:
![BA=\left[\begin{array}{cc}-2&1\\3&7\end{array}\right]\cdot \left[\begin{array}{cc}-2&4\\1&3\end{array}\right]=\left[\begin{array}{cc}-2\cdot (-2)+1\cdot 1&-2\cdot 4+1\cdot 3\\3\cdot (-2)+7\cdot 1&3\cdot 4+7\cdot 3\end{array}\right]=\left[\begin{array}{cc}5&-5\\1&33\end{array}\right]](https://tex.z-dn.net/?f=BA%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%261%5C%5C3%267%5Cend%7Barray%7D%5Cright%5D%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%264%5C%5C1%263%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%5Ccdot%20%28-2%29%2B1%5Ccdot%201%26-2%5Ccdot%204%2B1%5Ccdot%203%5C%5C3%5Ccdot%20%28-2%29%2B7%5Ccdot%201%263%5Ccdot%204%2B7%5Ccdot%203%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D5%26-5%5C%5C1%2633%5Cend%7Barray%7D%5Cright%5D)
C. Answers are not the same
D. Matrices multiplication is not commutastive in general, so
