All categories

2 years ago

What is the GCF of the terms of the polynomial −16y4 + 12y2 – 4y?

Mathematics

1 answer:

Tresset [83]2 years ago

5 0

Answer:

The GCF is - 4y

Step-by-step explanation:

Given the polynomial expression −16y4 + 12y2 – 4y?

If we carefully traverse all the terms we can factor out 4y from all the terms

=-4y(4y³-3y+1) which is similar to the original expression

Hence the greatest common factor GCF = - 4y

The Greatest Common Factor (GCF) is the largest number that will divided evenly into that number.

You might be interested in

If you were trying to decide whether to take out an auto loan for $6500 to buy your first car, thereby allowing you to commuto f

notka56 [123]

Answer:

Yes, a loan would meet our requirement to commute for an impressive summer internship program next year

Step-by-step explanation:

Taking a loan would meet our requirement of buying a car. We will be able to make the downpayment. This will enable us to buy a car. So the decision to take the loan will be valid. It will help us in commuting easily for the summer internship program. We will immediately get the car after making a down payment and will avail of the benefits of using the car. This is a healthy type of debt.

4 0

2 years ago

Equation: t = 25 + 8.25w Aaliyah needs $135.75 for a television. How many weeks will it take her to save enough money to buy a t

miss Akunina [59]

Not sure its totally right but can you grant me the most brainly?

5 0

1 year ago

Read 2 more answers

A population consists of 500 elements. We want to draw a simple random sample of 50 elements from this population. On the first

larisa [96]

Answer:

D. 0.002

Step-by-step explanation:

Given;

total number of sample, N = 500 elements

50 elements are to be drawn from this sample.

The probability of the first selection, out of the 50 elements to be drawn will be = 1 / total number of sample

The probability of the first selection = 1 / 500

The probability of the first selection = 0.002

Therefore, on the first selection, the probability of an element being selected is 0.002

The correct option is "D. 0.002"

4 0

2 years ago

Select all quantities that are proportional to the diagonal length of a square. A. Area of a square B. Perimeter of a square C.

viva [34]

Answer:

B.Perimeter of a square

6 0

1 year ago

Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang

kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

$P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164$

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

$P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145$

5 0

2 years ago