Let after n week both tanks have same quantity of water .
for first tank
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tank hold 540 gallon of water ,
leaking of water 4 gallon /week
so, in n week water leak from the tank =4× n = 4n
now, after n week quantity of water in the tank = 540 -4n
for second tank
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quantity of water in the tank =720 gallon
leaking of water at the rate = 7gallon/week
so, after n week water leak from the tank = 7× n =7n
after n week quantity of water rest in the tank = 720 -7n
now ,
according to question ,
water in first tank = water in second tank
540 -4n = 720 -7n
7n -4n = 720 -540
3n = 180
n = 60
hence , after 60 week both tank have same quantity of water.
If the width is 15, and the scale factor from width to length is 3:2, then you will have to divide 15 by 3 to get your original width.
15/3 = 5.
5 is your original width.
You take the decimal of say .48 and pull the it back so it would be 48%
Answer:
1yr 47,000, 2nd yr 48,880, 3rd yr 51,617.28, 4th yr 57,346.7981 or 57,346.80
Step-by-step explanation:
1yr 47,000
2nd yr 47,000 x 4% = 48,880
3rd yr 48,880 x 5.6% = 51,617.28
4th yr 51,617.28 x 11.1% = 57,346.7981 or 57,346.80
Answer:
![A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%26-4%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
message SHOW_ME_THE_MONEY_
Step-by-step explanation:
The matrix
![A=\left[\begin{array}{cc}1&4\\-1&-3\end{array}\right]\rightarrow |A|=(1 \times -3)-(-1\times 4)=1\\\rightarrow A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right] \\](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%264%5C%5C-1%26-3%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%20%7CA%7C%3D%281%20%5Ctimes%20-3%29-%28-1%5Ctimes%204%29%3D1%5C%5C%5Crightarrow%20A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%26-4%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
We can check that in fact A*A^⁻1=I_2 the identity matrix of size 2 x 2.
Now the message was divided in 1 x 2 matrices, then we have that the sequence given is the result of multiplying m by A, so to get m again we multiply now by A^⁻1. and we get the next table
Encoded message Decoded message message in letters by association
11 52 19 8 S H
-8 -9 15 23 O W
-13 -39 0 13 _ M
5 20 5 0 E _
12 56 20 8 T H
5 20 5 0 E _
-2 7 13 15 M O
9 41 14 5 N E
25 100 25 0 Y _
Then the message decoded is SHOW_ME_THE_MONEY_
