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3 years ago

The formula for the nth term of the arithmetic sequence first term is 9 common difference is -2

Mathematics

1 answer:

Yanka [14]3 years ago

3 0

Answer:

$a_n = 9 - 2(n-1)$

Step-by-step explanation:

Arithmetic sequence:

In an arithmetic sequence, the difference between consecutive terms is always the same, and is called common difference.

The nth term is given by the following equation:

$a_n = a_1 + (n-1)d$

In which $a_1$ is the first term and d is the common difference.

First term is 9 common difference is -2

This means that $a_1 = 9, d = -2$

So, the nth term is given by:

$a_n = a_1 + (n-1)d$

$a_n = 9 + (n-1)(-2)$

$a_n = 9 - 2(n-1)$

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3 years ago

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the theoretical probability that a man will have some form of color blindness is 8%,whereas the theoretical probability of a wom

lianna [129]

Answer:

If you have a group of 700 men, 56 will be color blind

Step-by-step explanation:

The theoretical probability that a man will have some form of color blindness is 8%.

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If you have a group of 700 men, 56 will be color blind

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3 years ago

Please Help!! Write the function I'm vertex form y = -x^2-2x+6 y=

balu736 [363]

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3 years ago

What is the following product? (2 square root 7+3 square root 6)(5 square root 2+4 square root 3)

Xelga [282]

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3 years ago

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What is the limit of f (×) as x approaches - infinty

tatyana61 [14]

If the degree of numerator and denominator are equal, then limit will be leading coefficient of numerator divided by the leading coefficient of denominator.

So then the limit would be 3/1 = 3.

Alternatively,

$\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}=\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}\cdot\dfrac{1/x^2}{1/x^2}=\lim_{x\to\infty}\dfrac{3+\frac6{x^2}}{1-\frac4{x^2}} = \dfrac{3+0}{1-0}=\boxed{3}$

Hope this helps.

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3 years ago