What is the recursive rule for the following arithmetic sequence? -7, -4, -1, 2, 5, …

Mathematics

1 answer:

Ludmilka [50]3 years ago

6 0

Answer:

Recursive rule for arithmetic sequence = an = a[n-1] + 3

Step-by-step explanation:

Given arithmetic sequence;

-7, -4, -1, 2, 5, …

Find:

Recursive rule for arithmetic sequence;

Computation:

Let a1 = -7

So,

⇒ a2 = a1 + 3 = -4

⇒ a3 = a2 + 3 = -1

⇒ a4 = a3 + 3 = 2

⇒ a5 = a4 + 3 = 5

So, the recursive formula is

⇒ an = a[n-1] + 3

Recursive rule for arithmetic sequence = an = a[n-1] + 3

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irakobra [83]

A= bh divide by 2 so you'll multiply 10x8 the product would be 80 and so you divide 80 by 2 and get 40 as your result.

6 0

3 years ago

Read 2 more answers

How do u do b??????????

horrorfan [7]

the assumption being that the first machine is the one on the left-hand-side and the second is the one on the right-hand-side.

the input goes to the 1st machine and the output of that goes to the 2nd machine.

if she uses and input of 6 on the 2nd one, the result will be 6² - 6 = 30, if we feed that to the 1st one the result will be √( 30 - 5) = √25 = 5, so, simply having the machines swap places will work to get a final output of 5.

clearly we can never get an output of -5 from a square root, however we can from the quadratic one, the 2nd machine/equation.

let's check something, we need a -5 on the 2nd, so

$\bf \underset{final~out put}{\stackrel{y}{-5}}=x^2-6\implies 1=x^2\implies \sqrt{1}=x\implies 1=x$

so if we use a "1" as the output on the first machine, we should be able to find out what input we need, let's do that.

$\bf \underset{first~out put}{\stackrel{y}{1}}=\sqrt{x-5}\implies 1^2=x-5\implies 1=x-5\implies 6=x$

so if we use an input of 6 on the first machine, we should be able to get a -5 as final output from the 2nd machine.

$\bf \stackrel{first~machine}{y=\sqrt{\boxed{6}-5}}\implies y=\sqrt{1}\implies y=1 \\\\\\ \stackrel{second~machine}{y = \boxed{1}^2-6}\implies y = 1-6\implies y = -5$