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3 years ago

Factor the polynomial 54c3d4 + 9c4d2

Mathematics

2 answers:

miv72 [106K]3 years ago

7 0

54c^3d^4 + 9c^4d^2
9c^3d^2(6d^2 + c)

Harlamova29_29 [7]3 years ago

3 0

Answer: $9c^3d^2(6d^2+c)$

Step-by-step explanation:

Given expression : $54c^3d^4+9c^4d^2$

We can write the above expression as

$(6\times9)c^3d^{(2+2)}+9c^{(3+1)}d^2$

$=(6\times9)c^3d^2d^2+9c^3c^1d^2$

We can see the greatest common factor of both the terms = $9c^3d^2$

Taking the greatest common factor outside , we get

$9c^3d^2(6d^2+c)$

Hence, the factorize form of given expression is given by:-

$54c^3d^4+9c^4d^2=9c^3d^2(6d^2+c)$

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1. A fish is 12 meters below the surface of the ocean. What is its elevation? meters 2. A sea bird is 28 meters above the surfac

Vlad1618 [11]

Answer:

- 12 meters

28 meters

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Step-by-step explanation:

Given that:

A fish is 12 meters below the surface of the ocean. What is its elevation?

Position of fish = 12 meters below the surface = - 12 ;

Sea level = 0

Elevation = sea level + position

Elevation = 0 + (- 12)

= - 12 meters

2. A sea bird is 28 meters above the surface of the ocean. What is its elevation?

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Sea level = 0

Elevation = sea level + position

Elevation = 0 + 28

= 28 meters

3. If the bird is directly above the fish, how far apart are they?

If bird is directly above the fish;

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28 - (-12)

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3 years ago

1. (k+7)² =289 2. (2s-1)² =225 3. (x-4)² =169 please help me answer these :)

professor190 [17]

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k=10 \vee k=-24$

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2s-1=15\vee 2s-1=-15\\
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Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

DENIUS [597]

Answer:

The probability that a randomly chosen plate contains the number 2222 is 0.000028 approximately.

The probability that a randomly chosen plate contains the sub-string HI is 0.002548 approximately.

Step-by-step explanation:

Consider the provided information.

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

Part (A)

Let A is the ways in which plates consist of three letters followed by three digits and B is the ways in which two letters followed by four digits.

Here repetition is allow. The number of alphabets are 26 and the number of distinct digits are 10.

The numbers of ways in which three letters followed by three digits can be chosen is: $26\times 26\times 26 \times 10 \times 10 \times10$

$26^3\times 10^3=17576000$

The numbers of ways in which two letters followed by four digits can be chosen is: $26\times 26\times 10 \times 10 \times 10 \times10$

$26^2\times 10^4=6760000$

Hence, the total number of ways are 17576000 + 6760000 = 24336000

Randomly chosen plate contains the number 2222 that means the first two letter can be any alphabets but the rest of the digit should be 2222.

Thus, the total number of ways that a randomly chosen plate contains the number 2222 number are: $26^2=676$

The probability that a randomly chosen plate contains the number 2222 is:

$\frac{676}{24336000} \approx 0.000028$

Part (B)

The number of ways in which chosen plate contains the sub-string HI:

If three letters followed by three digits plate contains the sub-string HI, then the number of possible ways are:

$26\times 1\times10^3+1\times 26\times10^3$

If two letters followed by four digits plate contains the sub-string HI, then the number of possible ways are:

$1\times 10^4$

Thus, the total number of ways that a randomly chosen plate contains the sub-string HI are:

$1\times 10^4+26\times 1\times10^3+1\times 26\times10^3$

$62000$

From part (A) we know that the total number of ways to chose a number plate is 24336000.

The probability that a randomly chosen plate contains the sub-string HI is:

$\frac{62000}{24336000} \approx 0.002548$

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