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09.07.2019 •
Mathematics
Will give brainliest if right! the e. coli bacterium is about 5×10^−7 meters wide. a hair is about 1.7×10^−5 meters wide. which is wider, the bacterium or the hair? the bacterium the same width the hair
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Ответ:
1.7 × 10^−5 = 0.000017
answer
The bacterium is wider
Ответ:
d. 12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + 5x^55)
Step-by-step explanation:
First, we find the greatest common factor of 12x^23y^40z^9 and 10x^78z^5.
We work on the numbers, then on x, then on y, then on z.
We start with the GCF of 12 and 10.
10 = 2 * 5
12 = 2 * 3 * 3
The only factor 10 and 12 have in common is 2, so the GCF of 12 and 10 is 2.
Now we work on x. We have x^23 and x^78.
x^23 has 23 factors of x.
x^78 has 78 factors of x.
Since x^23 and x^78 have both at least 23 factors of x, x^23 is the GCF of x.
Now we work on y.
There is y^40 in one term, but there is no y in the other term. There is no GCF for the y.
Now we work on z. The first term has z^9. the second term has z^5. They both have at least 5 factors of z, so the GCF of z is z^5.
Now we put all the parts of the GCF together, and the GCF of the two terms is
2x^23z^5
We use option d. to factor.
12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(___a + b___)
We already have the GCF outside the parentheses. Now we need to figure out what goes inside the parentheses.
The terms that go in positions a and b are the terms that when multiplied by the GCF give you the original expression.
We now work on position a.
What do you multiply by 2x^23z^5 to get 12x^23y^40z^9?
6y^40z^4
Now we have
12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + b___)
Now we do the same for position b.
What do you multiply by 2x^23z^5 to get 10x^78z^5?
5x^55
Now we have
12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + 5x^55)
d. 12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + 5x^55)