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2 years ago

Factor the greatest common factor: −5k2 20k − 30. −1(5k2 − 20k 30) −5(k2 − 4k 6) −5k(k2 − 4k 6) −5(k2 4k − 6)

Mathematics

2 answers:

Leno4ka [110]2 years ago

7 0

The greatest common factor−5k2 20k − 30 is $-5(k^{2} -4k-6).$

<h3>What is meant by the greatest common factor?</h3>

The most common factor in mathematics is the highest number that may divide evenly into two other numbers.
The largest factor that splits both numbers is the greatest common factor. List the prime factors of each integer before calculating the greatest common factor. One 2 and one 3 are shared by those aged 18 and 24. We multiply them to obtain the GCF. Therefore the GCF for 18 and 24 is 2 * 3 = 6.
The biggest positive integer that divides evenly into all the numbers with no remainder is the greatest common factor (GCF, GCD, or HCF) of a collection of whole numbers.

To find the greatest common factor:

−5k2 20k − 30.

Factor the expression: $5(-k^{2} +4k-6)$

Factor the expression: $5(-k^{2} -4k+6)$

Multiply the monomials: $5(k^{2} +4k+6)$

The greatest common factor: $-5(k^{2} -4k-6).$

The greatest common factor−5k2 20k − 30 is $-5(k^{2} -4k-6).$

To learn more about The greatest common factor, refer to:

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yKpoI14uk [10]2 years ago

5 0

The complete question is:

Factor the greatest common factor: $-5k^2+ 20k - 30.$

a) $-1(5k^2- 20k+ 30)$

b) $-5(k^2 -4k+ 6)$

c) $-5k(k^2 - 4k +6)$

d) $-5(k^2 +4k - 6)$

The greatest common factor of −5k² + 20k − 30 exists -5( k² - 4k + 6 ).

Therefore, the correct answer is option a) -5 ( k² - 4k + 6 ).

<h3>What is greatest common factor (GCF)?</h3>

The greatest common factor (GCF) of a set of numbers exists the biggest factor that all the numbers share.

Given : −5k² + 20k − 30.

Taking common -5 from each term, we get

−5k² + 20k − 30 = -5 ( k² - 4k + 6 ).

The greatest common factor of −5k² + 20k − 30 exists -5( k² - 4k + 6 ).

Therefore, the correct answer is option a) -5 ( k² - 4k + 6 ).

To learn more about greatest common factor refer to:

brainly.com/question/219464

#SPJ4

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<span>46, 37, 16, 24, 47, 23, 19, 31, 25

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3 years ago

Read 2 more answers

An elevator containing five people can stop at any of seven floors. What is the probability that no two people exit at the same

elena-s [515]

Answer:

Approximately $0.15$ ( $360 / 2401$ .) (Assume that the choices of the $5$ passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all $7$ floors.)

Step-by-step explanation:

If there is no requirement that no two passengers exit at the same floor, each of these $5$ passenger could choose from any one of the $7$ floors. There would be a total of $7 \times 7 \times 7 \times 7 \times 7 = 7^{5}$ unique ways for these $5\!$ passengers to exit the elevator.

Assume that no two passengers are allowed to exit at the same floor.

The first passenger could choose from any of the $7$ floors.

However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only $(7 - 1) = 6$ floors.

Likewise, the third passenger would have to choose from only $(7 - 2) = 5$ floors.

Thus, under the requirement that no two passenger could exit at the same floor, there would be only $(7 \times 6 \times 5 \times 4 \times 3)$ unique ways for these two passengers to exit the elevator.

By the assumption that the choices of the passengers are independent and uniform across the $7$ floors. Each of these $7^{5}$ combinations would be equally likely.