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marin [14]
3 years ago
11

Write $6.45$ as a reduced fraction.

Mathematics
1 answer:
denis-greek [22]3 years ago
5 0

Do you want to know how to write the decimal number 6.45 as a fraction?

Here we will show you step-by-step how to convert 6.45 so you can write it as a fraction.

You can take any number, such as 6.45, and write a 1 as the denominator to make it a fraction and keep the same value, like this:

6.45 / 1

To get rid of the decimal point in the numerator, we count the numbers after the decimal in 6.45, and multiply the numerator and denominator by 10 if it is 1 number, 100 if it is 2 numbers, 1000 if it is 3 numbers, and so on.

Therefore, in this case we multiply the numerator and denominator by 100 to get the following fraction:

645 / 100

Then, we need to divide the numerator and denominator by the greatest common divisor (GCD) to simplify the fraction.

The GCD of 645 and 100 is 5. When we divide the numerator and denominator by 5, we get the following:

129 / 20

Therefore, 6.45 as a fraction is as follows:

129 / 20

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The answer would be 4/8 and 7/8
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7 0
3 years ago
Please help ASAP!!!!!!
erica [24]

Answer:

300

Step-by-step explanation:

20*30=600

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6 0
4 years ago
The coordinates J(2, 2), K(-1, 3), L(-2, -1) form a right triangle. True or False.
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7 0
3 years ago
Read 2 more answers
Which expression is equivalent to (8x^3)^2/3
VARVARA [1.3K]

Answer: 64x^6/3

Step-by-step explanation:

(8x^3)^2/3

8^2(x^3)^3/3

64(x^3)^3/3

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Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great Superbowl Weekend! :-)

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4 0
3 years ago
N is the sum of three consecutive primes and is also the product of two 2-digit primes, what is the least possible value of N
Nookie1986 [14]

Hello from MrBillDoesMath!

Answer:   N = 143

Discussion:

This one took some trial and error! At first I listed all 2 digit primes, looked at the list, but didn't know how to proceed. So, I took the smallest 2 digit primes numbers:  11 and 13 and wondered if their product, 13*11 = 143, could be represented as the sum of  3 consecutive primes.   I went back to my list of primes,  added groups of three consecutive numbers that seemed to be in the right range to give the desired sum, and stumbled on  43, 47, and 53!

43 + 47 + 53 =  143 !


Therefore N = 143. It's the sum of 43, 47, and 53 as well as the product of 11 and 13.


Thank you,

MrB


7 0
3 years ago
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