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4 years ago

What is the equation of an arithmetic sequence with a first term of 6 and a second term of 2?

Mathematics

1 answer:

kifflom [539]4 years ago

3 0

Answer:

The equation of the sequence = 10 - 4n

Step-by-step explanation:

The rule of the arithmetic sequence is Tn = a + (n - 1)d

Where a is the first term , d is the difference between each consecutive terms and n is the position of the number in the sequence

∵ First term = 6

∵ Second term = 2

∴ The difference between the second and the first = 2 - 6 = -4

∴ Tn = 6 + (n - 1)(-4) = 6 - 4n + 4 = 10 - 4n

∴ The equation of the sequence = 10 - 4n

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What is the correct order of operations for simplifying the expression (x+3)^2-(x^2+9)/2x^2?

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Answer:

The given expression is simplified as $\frac{(x+3)^2-(x^2 +9)} {2x^2} = \frac{3}{x}$

Step-by-step explanation:

Here, the given expression is:

$\frac{(x+3)^2-(x^2 +9)} {2x^2}$

Now, with ALGEBRAIC IDENTITIES:

$(a+b)^2 = a^2 + b^2 + 2ab$

Now, similarly: $(x+3)^2 = x^2 + (3)^2 + 2x(3) = (x^2 + 9 + 6x)$

Now, substituting the values in the given expression, we get:

$\frac{(x+3)^2-(x^2 +9)} {2x^2} \implies \frac{(x^2 + 9 + 6x)-(x^2 +9)} {2x^2}\\= \frac{(x^2 + 9 + 6x-x^2 -9)} {2x^2} \\= \frac{ 6x} {2x^2} = \frac{3}{x}$

Hence, the given expression $\frac{(x+3)^2-(x^2 +9)} {2x^2} = \frac{3}{x}$

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