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Katen [24]
3 years ago
7

12a^3b^2 +18a²b^2 – 12ab^2 Factor completely

Mathematics
1 answer:
AleksAgata [21]3 years ago
4 0

The factorization of 12a^3b^2 +18a²b^2 – 12ab^2 is 6 a b^{2}(a+2)(2 a-1)

<u>Solution:</u>

Given, expression is 12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}

We have to factorize the given expression completely.

Now, take the expression

12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}

Taking b^2 as common term,

b^{2}\left(12 a^{3}+18 a^{2}-12 a\right)

Taking "a" as common term,

b^{2}\left(a\left(12 a^{2}+18 a-12\right)\right)

Taking "6" as common term,

b^{2}\left(a\left(6\left(2 a^{2}+3 a-2\right)\right)\right)

Splitting "3a" as "4a - a" we get,

b^{2}\left(a\left(6\left(2 a^{2}+4 a-a-2\right)\right)\right)

\begin{array}{l}{b^{2}(a(6(2 a(a+2)-1(a+2))))} \\\\ {b^{2}(a(6((a+2) \times(2 a-1))))} \\\\ {6 a b^{2}(a+2)(2 a-1)}\end{array}

Hence, the factored form of given expression is 6 a b^{2}(a+2)(2 a-1)

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Noah and Lin are making paper cones to hold popcorn to hand out at parent math night. They want the cones to hold 9 cubic inches
DanielleElmas [232]

Answer:

So, The possible values of h and r are

1. r = 3 , h = 3

2. r = 1, h = 27

3. r = 2, h = 6.75

Step-by-step explanation:

P.S - The exact ques is -

As we know that Volume of cone , V = \frac{1}{3}\pi r^{2}h

As given,

V = 9\pi

⇒\frac{1}{3}\pi r^{2}h = 9\pi \\  \frac{1}{3} r^{2}h = 9 \\r^{2}h = 27

So,

The possible values of h and r are

1. r = 3 , h = 3

2. r = 1, h = 27

3. r = 2, h = 6.75

In Case I-

When r = 3, h = 3

(3)² ×3 = 9×3 = 27

Satisfied

Case II-

r = 1, h = 27

(1)²×27 = 1×27 = 27

Satisfied

Case III-

r = 2, h = 6.75

(2)²×6.75 = 4×6.75 = 27

Satisfied

8 0
3 years ago
1Y, the number of accidents per year at a given intersection, is assumed to have a Poisson distribution. Over the past few years
miss Akunina [59]

Answer:

The probability that the intersection will come under the emergency program is 0.1587.

Step-by-step explanation:

Lets divide the problem in months rather than in years, because it is more suitable to divide the period to make a better approximation. If there were 36 accidents in average per year, then there should be 3 accidents per month in average. We can give for the amount of accidents each month a Possion distribution with mean 3 and variance 3.

Since we want to observe what happen in a period of one year, we will use a sample of 12 months and we will take its mean. We need, in average, more than 45/12 = 3.75 accidents per month to confirm that the intersection will come under the emergency program.

For the central Limit theorem, the sample mean will have a distribution Normal with mean 3 and variance 3/12 = 0.25; thus its standard deviation is √0.25 = 1/2.

Lets call the sample mean distribution X. We can standarize X obtaining a standard Normal random variable W with distribution N(0,1).

W = \frac{X-\mu}{\sigma} = \frac{X-3}{1/2} = 2x-6

The values of \phi , the cummulative distribution function of W, can be found in the attached file. We are now ready to compute the probability of X being greater than 3.75, or equivalently, the probability than in a given year the amount of accidents is greater than 45, leading the intersection into an emergency program

P(X > 3.75) = P(2X-6 > 2*3.75-6) = P(W > 1) = 1-\phi(1) = 1-0.8413 \\= 0.1587

Download pdf
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Answer:

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3 years ago
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this would be the answer because if you take a positive and add it to a negative you get zero if their the additive inverse of the number.

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