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Ira Lisetskai [31]
3 years ago
13

How do you Divide 61,986 ÷23

Mathematics
1 answer:
Minchanka [31]3 years ago
6 0

Answer:

2695.04348

Well, I don't want to do ALL of the work but I'll give you an idea of how I got it:

First, ask yourself, "How many times does 23 go into 61?" which is the same thing as 61 divided by 23.

23 times 2 equals 46. I chose this number because 23 times 3 equals 69 and that is too much.

(The first digit is 2)

So 61 minus 46 equals 15. Then you bring down the 9 and divide 159 by 23.

23 times 6 is 138 and the closest you will get. Again, 159 minus 138 equals 21.

(Now the second digit of the answer is 6)

Bring down the 8 to get 218. 218 divided by 23 equals 9 (well, not exactly, but you get what I mean). 23 times 9 equals 207. 218 minus 207 equals 11.

(Don't forget, the third digit in the answer is now 9)

Keep doing this until you get your final answer, which I already gave you at the top to double check. When you reach the last number of 61,986 (which is 6), go into the decimal place and keep on working from there.

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Solve the equation (linear equation)<br><br> <img src="https://tex.z-dn.net/?f=8%5E%7B2x%2B7%7D%20%3D%20%28%5Cfrac%7B1%7D%7B32%7
Minchanka [31]

Answer: x=-1

Step-by-step explanation:

By the negative exponent rule, you have that:

(\frac{1}{a})^n=a^{-n}

By the exponents properties, you know that:

(m^n)^l=m^{(nl)}

Therefore, you can rewrite the left side of the equation has following:

(\frac{1}{8})^{-(2x+7)}=(\frac{1}{32})^{3x}

 Descompose 32 and 8 into its prime factors:

32=2*2*2*2*2=2^5\\8=2*2*2=2^3

Rewrite:

(\frac{1}{2^3})^{-(2x+7)}=(\frac{1}{2^5})^{3x}

Then:

(\frac{1}{2})^{-3(2x+7)}=(\frac{1}{2})^{5(3x)}

As the base are equal, then:

-3(2x+7)=5(3x)

Solve for x:

-6x-21=15x\\-21=15x+6x\\-21=21x\\x=-1

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