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2 years ago

Explain the difference between an arithmetic and geometric sequence

Mathematics

1 answer:

Feliz [49]2 years ago

3 0

Answer:

Arithmetic sequences have a constant difference between consecutive terms. On the other hand, geometric sequences have a constant raito. In an arithmetic sequence, the terms can be determined by adding or subtracting a number to the preceding term, while for geometric sequences, each term is determined by multiplying or dividing a number to the preceding term.

Step-by-step explanation:

Explanation above.

hope this helps :)

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6.089 x 3.4 give explanation

kati45 [8]

Answer:

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Step-by-step explanation:

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2 years ago

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Simplify; 1/x^2+5x+6 + 1/x^2+3x+2 pls I need a step by step solution urgently

ikadub [295]

Answer:

The simplest form of $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ is $\frac{2}{(x+1)(x+3)}$

Step-by-step explanation:

To add two algebraic fractions do these steps

Factorize each denominator
Simplify each fraction to its lowest term
Find the LCM of the two denominators
Divide LCM by each denominator and multiply the numerators by its corresponding quotients
Add the two fraction and simplify the last answer

To simplify $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$

Factorize each denominator

∵ The denominator of the first fraction is x² + 5x + 6

∵ x² = (x)(x)

∵ 6 = (2)(3)

∵ (2)(x) + (3)(x) = 5x ⇒ middle term

∴ x² + 5x + 6 = (x + 2)(x + 3)

∵ The denominator of the second fraction is x² + 3x + 2

∵ x² = (x)(x)

∵ 2 = (2)(1)

∵ (2)(x) + (1)(x) = 3x ⇒ middle term

∴ x² + 3x + 2 = (x + 2)(x + 1)

Find the LCM of the two denominators

∵ The denominators are (x + 2)(x + 3) and (x + 2)(x + 1)

- LCM is all the different factors multiplied together

∴ LCM of them is (x + 1)(x + 2)(x + 3)

- Divide LCM by each denominator

∵ (x + 1)(x + 2)(x + 3) ÷ (x + 2)(x + 3) = (x + 1)

- Multiply the numerator of the first fraction by (x + 1)

∵ (x + 1)(x + 2)(x + 3) ÷ (x + 2)(x + 1) = (x + 3)

- Multiply the numerator of the second fraction by (x + 3)

∴ $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ = $\frac{x+1}{(x+1)(x+2)(x+3)}+\frac{x+3}{(x+1)(x+2)(x+3)}$

- Add the numerators and write the answer as a single fraction

∴ $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ = $\frac{2x+4}{(x+1)(x+2)(x+3)}$

- Factorize the numerator by taking 2 as a common factor

∵ 2x + 4 = 2(x + 2)

∴ $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ = $\frac{2(x+2)}{(x+1)(x+2)(x+3)}$

- Simplify the fraction by dividing up and down by (x + 2)

∴ $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ = $\frac{2}{(x+1)(x+3)}$

The simplest form of $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ is $\frac{2}{(x+1)(x+3)}$

5 0

3 years ago

Consider the system of differential equations dxdt=−4ydydt=−4x. Convert this system to a second order differential equation in y

koban [17]

$\dfrac{\mathrm dy}{\mathrm dt}=-4x\implies x=-\dfrac14\dfrac{\mathrm dy}{\mathrm dt}\implies\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}$