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3 years ago

Find the product of (x2 + 3x + 1)(x2 + x + 2)

Mathematics

2 answers:

Elza [17]3 years ago

7 0

<h2>Hello!</h2>

The answer is: $x^{4}+4x^{3}+6x^{2}+7x+2$

To find the product of the given expression, we need to apply the distributive property and then, group each like term.

The distributive property states that:

(a+b+c)(d+e+f)=a*d+a*e+a*f+b*d+b*e+b*f+c*d+c*e+c*f

Then, we have to group each like term. Like terms are the terms that have the same variable and exponent , for example:

$x+x+x^{2}=2x+x^{2}$

For this case, the variable "x" has two Like terms, since both have the same exponent (1), then, we add each other in order to group term.

Also, we must remember the following power property:

Product of powers property,

$a^{n}*a^{m}=a^{n+m}$

So, solving the product we have:

$(x^{2}+3x+1)(x^{2}+x+2)=x^{2}*x^{2}+x*x^{2}+2x^{2}+3x*x^{2}+3x*x+6x+x^{2}+x+2\\(x^{2}+3x+1)(x^{2}+x+2)=x^{2+2}+x^{1+2}+2x^{2}+3x^{2+1}+3x^{1+1}+6x+x^{2}+x+2\\(x^{2}+3x+1)(x^{2}+x+2)=x^{4}+4x^{3}+6x^{2}+7x+2$

Have a nice day!

Bogdan [553]3 years ago

4 0

Answer: $x^4+4x^3+6x^2+7x+2$

Step-by-step explanation:

To solve the exercise you must apply the following proccedure:

- Apply the distributive property.

- Keep on mind that when you multiply two powers with equal base, you must add the exponents.

- Add like terms.

Therefore, you obtain the following product:

$=x^{(2+2)}+x^{(2+1)}+2x^2+3x^{(1+2)}+3x^{(1+1)}}+6x+x^{2}+x+2\\=x^4+x^3+2x^2+3x^3+3x^2+6x+x^2+x+2\\=x^4+4x^3+6x^2+7x+2$

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a company pays its salespeople $2,000 a month and a commission of 18% of their total sales (T). which of the following equations

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Answer:

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18% of 2,000 is 360

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3 years ago

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4 years ago

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3 years ago

Please help very urgent

Alik [6]

Answer:

32, <u>16</u>, <u>8</u>, <u>4</u>, <u>2</u>, 1

Explanation:

The geometric mean can be represented by $\sqrt[n]{x_{1} • x_{2} • x_{3} • .. x_{n}}$ .

Which is the mean of the product of n numbers, used to find the average of a geometric progression.

Don't get confused by geometric mean, it is only asking you about the next numbers in the geometric sequence given the first and sixth term.

The explicit rule for a geometric sequence can be modeled by:

$a_{n} = a_{1} • r^{n-1}$

Where $a_{n}$ is the nth term, $a_{1}$ is the first term in the sequence, n is the term number, and r is the common ratio.

Since we already know the first term, $a_{1}$ will simply be 32.

Since we know it's geometric, there will be an exponential relationship, which means that we will use the geometric mean to find the common ratio.

There are 6 total terms, r is raised to the n – 1 so 6 – 1 = 5, and that will be the degree of this root.

$\sqrt[5]{\frac{a_{6}}{a_{1}}}$ =

$\sqrt[5]{\frac{1}{32}}$ =

$\frac{1}{2}$ .

Therefore: $r = \frac{1}{2}$ .

Using all the information we have, we can find the explicit rule:

$a_{n} = a_{1} • r^{n-1}$

$a_{1}$ = 32
$r = \frac{1}{2}$

$a_{n} = a_{1} • r^{n-1}$ →

$\boxed{a_{n} = 32 • (\frac{1}{2})^{n-1}}$

________________________________

We can test that this works by substituting the number location of the term you want to find.

For instance:

$a_{1} = 32 • (\frac{1}{2})^{1-1}$

$a_{1} = 32 • (\frac{1}{2})^{0}$

$a_{1} = 32 • 1$

$a_{1} = 32$

$a_{6} = 32 • (\frac{1}{2})^{6-1}$

$a_{6} = 32 • (\frac{1}{2})^{5}$

$a_{6} = 32 • \frac{1}{32}$

$a_{6} = 1$

3 0

3 years ago

Read 2 more answers

A marine aquarium has a small tank and a large tank, each containing only red and blue fish. In each tank, the ratio of red fish

Anna007 [38]

Answer:

Ratio of blue fish in the small tank to the red fish in large tank is 10 : 6279

Step-by-step explanation:

Let the number of red fish and blue fish in the large tank are x and y respectively.

Similarly ratio of red fish and blue fish in the small tank are x' and y' respectively.

Since in each tank ratio of the red fish to blue fish is 333 : 444

That means x : y = 333 : 444

Or $\frac{x}{y}=\frac{333}{444}$

⇒ $\frac{x}{y}=\frac{3}{4}$

⇒ y = $\frac{4x}{3}$ --------(1)

Similarly x' : y' = 333 : 444

⇒ $\frac{x'}{y'}=\frac{333}{444}$

⇒ $\frac{x'}{y'}=\frac{3}{4}$

⇒ x' = $\frac{3y'}{4}$ ------(2)

Ratio of the fish in large tank to the fish in small tank is 464646 : 555

So (x + y) : (x' + y') = 464646 : 555

$\frac{x+y}{x'+y'}=\frac{464646}{555}$

Now we replace the values of x and y' from equation (1) and equation (2)

$\frac{(x+\frac{4x}{3})}{(\frac{3y'}{4}+y')}=\frac{464646}{555}$

$\frac{\frac{7x}{3}}{\frac{7y'}{4}}=\frac{464646}{555}$

$\frac{4}{3}\times \frac{x}{y'}=\frac{464646}{555}$

$\frac{x}{y'}=\frac{3}{4}\times \frac{464646}{555}$

$\frac{x}{y'}=\frac{232323}{370}$

$\frac{y'}{x}=\frac{370}{232323}$

$\frac{y'}{x}=\frac{10}{6279}$

Therefore, ratio of blue fish in the small tank to the red fish in large tank is 10 : 6279

4 0

4 years ago