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larisa [96]
3 years ago
8

Need help Asap thank you in advance

Mathematics
1 answer:
julia-pushkina [17]3 years ago
3 0

Answer:

12x^4-20x^2+32x=4x(3x^3-5x+8)

Step-by-step explanation:

<u>Common Factors</u>

An algebraic expression that is formed by sums or subtractions of terms can be factored provided there are numeric or variable common factors in all the terms.

The following expression

Z=12x^4-20x^2+32x

Can be factored in the constants and in the variable x.

1. To find the common factor of the variable, we must locate if the variable is present in all terms. If so, we take the common factor as the variable with an exponent which is the lowest of all the exponents found throughout the different terms. In this case, the lowest exponent is x (exponent 1).

2. To find the common factors of the constants, we take all the coefficients:

12 - 20 - 32

and find the greatest common divisor of them, i.e. the greatest number all the given numbers can be divided by. This number is 4, since 12/4=3, 20/4=5 and 32/4=8

3. The factored expression is

Z=12x^4-20x^2+32x=4x(3x^3-5x+8)

\boxed{12x^4-20x^2+32x=4x(3x^3-5x+8)}

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This math my graduation depends on it
In-s [12.5K]

In an arithmetic sequence, consecutive terms have a fixed distance d between them. If a₁ is the first term, then

2nd term = a₂ = a₁ + d

3rd term = a₃ = a₂ + d = a₁ + 2d

4th term = a₄ = a₃ + d = a₁ + 3d

and so on, up to

nth term = a_n = a_{n-1} + d = a_{n-2} + 2d = a_{n-3} + 3d = \cdots = a_1 + (n-1)d

so that every term in the sequence can be expressed in terms of a₁ and d.

6. It's kind of hard to tell, but it looks like you're given a₁₃ = -53 and a₃₅ = -163.

We have

a₁₃ = a₁ + 12d = -53

a₃₅ = a₁ + 34d = -163

Solve for a₁ and d. Eliminating a₁ and solving for d gives

(a₁ + 12d) - (a₁ + 34d) = -53 - (-163)

-22d = 110

d = -5

and solving for a₁, we get

a₁ + 12•(-5) = -53

a₁ - 60 = -53

a₁ = 7

Then the nth term is recursively given by

a_n = a_{n-1}-5

and explicitly by

a_n = 7 + (n-1)(-5) = 12 - 5n

7. We do the same thing here. Use the known terms to find a₁ and d :

a₁₉ = a₁ + 18d = 15

a₃₈ = a₁ + 37d = 72

⇒   (a₁ + 18d) - (a₁ + 37d) = 15 - 72

⇒   -19d = -57

⇒   d = 3

⇒   a₁ + 18•3 = 15

⇒   a₁ = -39

Then the nth term is recursively obtained by

a_n = a_{n-1}+3

and explicitly by

a_n = -39 + (n-1)\cdot3 = 3n-42

8. I won't both reproducing the info I included in my answer to your other question about geometric sequences.

We're given that the 1st term is 3 and the 2nd term is 12, so the ratio is r = 12/3 = 4.

Then the next three terms in the sequence are

192 • 4 = 768

768 • 4 = 3072

3072 • 4 = 12,288

The recursive rule with a₁ = 3 and r = 4 is

a_n = 4a_{n-1}

and the explicit rule would be

a_n = 3\cdot4^{n-1}

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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