Answer:
Left side
Ca(OH)2 --> Ca + 2 O + 2 H
2 HBr --> 2 H + 2 Br
Together: Ca + 2 O + 4 H + 2 Br
You have only 1 Br and only 3 H. So you must correct this.
It can expand by 48600.
By depositing 5400 this represents the 10 percent that needs to be reserved. multiply 5400 by 10 to get 100 percent that could be loaned or created. 54000 subtract the initial 10 percent of 5400 to get 48600 difference or new money that could be created.
Using the powers of 9, it is found that the last digit of the product will always be either 1 or 9.
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The multiplication of as many 9's as you want is equivalent to the powers of 9, which are:
![9^2 = 9 \times 9 = 81](https://tex.z-dn.net/?f=9%5E2%20%3D%209%20%5Ctimes%209%20%3D%2081)
![9^3 = 9 \times 9 \times 9 = 729](https://tex.z-dn.net/?f=9%5E3%20%3D%209%20%5Ctimes%209%20%5Ctimes%209%20%3D%20729)
![9^4 = 9 \times 9 \times 9 \times 9 = 6561](https://tex.z-dn.net/?f=9%5E4%20%3D%209%20%5Ctimes%209%20%5Ctimes%209%20%5Ctimes%209%20%3D%206561)
![9^5 = 9 \times 9 \times 9 \times 9 \times 9= 59049](https://tex.z-dn.net/?f=9%5E5%20%3D%209%20%5Ctimes%209%20%5Ctimes%209%20%5Ctimes%209%20%5Ctimes%209%3D%2059049)
....
![9^{10} = 3486784401](https://tex.z-dn.net/?f=9%5E%7B10%7D%20%3D%203486784401)
Thus, the last digit of the products will always be either 1 or 9.
A similar problem is given at brainly.com/question/20692373
Answer:
Yes!
Step-by-step explanation:
Lets say you have 4 numbers. (1-4)
Draw a box and separate the box into 4's (1 for each number)
On 2 sides of the whole box write the numbers (1 on one side, 2 on the same side but below it, 3 on the top of the box, 4 besides the other number)
After setting it up, you multiply them just like a times table.