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salantis [7]
3 years ago
11

There are 3 teams of 18 employees working today. Company

Mathematics
1 answer:
Degger [83]3 years ago
5 0
First do 3 × 18 because there are 3 different teams and each one has 18 workers. 3 × 18 = 54. So you only need 9 people on the floor. 9/54 in fraction form. Are there a certain number of floors? I think this is the answer to the question you have written question though.
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What is the factorization of 729^15+1000?
igomit [66]

Answer:

The factorization of 729x^{15} +1000 is (9x^{5} +10)(81x^{10} -90x^{5} +100)

Step-by-step explanation:

This is a case of factorization by <em>sum and difference of cubes</em>, this type of factorization applies only in binomials of the form (a^{3} +b^{3} ) or (a^{3} -b^{3}). It is easy to recognize because the coefficients of the terms are <u><em>perfect cube numbers</em></u> (which means numbers that have exact cubic root, such as 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, etc.) and the exponents of the letters a and b are multiples of three (such as 3, 6, 9, 12, 15, 18, etc.).

Let's solve the factorization of 729x^{15} +1000 by using the <em>sum and difference of cubes </em>factorization.

1.) We calculate the cubic root of each term in the equation 729x^{15} +1000, and the exponent of the letter x is divided by 3.

\sqrt[3]{729x^{15}} =9x^{5}

1000=10^{3} then \sqrt[3]{10^{3}} =10

So, we got that

729x^{15} +1000=(9x^{5})^{3} + (10)^{3} which has the form of (a^{3} +b^{3} ) which means is a <em>sum of cubes.</em>

<em>Sum of cubes</em>

(a^{3} +b^{3} )=(a+b)(a^{2} -ab+b^{2})

with a= 9x^{5} y b=10

2.) Solving the sum of cubes.

(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)((9x^{5})^{2}-(9x^{5})(10)+10^{2} )

(9x^{5})^{3} + (10)^{3}=(9x^{5} +10)(81x^{10}-90x^{5}+100)

.

8 0
2 years ago
Help please<br> Calculate the product of 8/15, 6/5, and 1/3. 
bogdanovich [222]

Hey there!

Great question:)


8/15 * 6/5 * 1/3 = 16/75


I hope this helps;)

4 0
3 years ago
Suppose that a, b \in \mathbb{Z}a,b∈Z, not both 00, and let d=\gcd(a, b)d=gcd(a,b). Bezout's theorem states that dd can be writt
lara [203]

Answer:

Step-by-step explanation:

Recall that we say that d | a if there exists an integer k for which a = dk. So, let d = gcd(a,b) and let x, y be integers. Let t = ax+by.

We know that d | a, d | b so there exists integers k,m such that a = kd and b = md. Then,

t = ax+by = (kd)x+(md)y = d(kx+my). Recall that since k,  x, m, y are integers, then (kx+my) is also an integer. This proves that d | t.

3 0
3 years ago
Rewrite in simplest terms: -4(5v – 3v – 6) – 9v
julsineya [31]

Answer:

-4 (5v -3v -6) -9v (use the distributive property)

-20v +12v +24 -9v (combine all like terms)

-17v +24

3 0
2 years ago
Read 2 more answers
Help, please :)<br> appreciate it
myrzilka [38]

27). 8a+7

28). 10k-1

29). 2n-48

30). x+16

31). 14b+3c-3

32) -2y=10

6 0
3 years ago
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