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3 years ago

11

There are 3 teams of 18 employees working today. Company

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

3 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