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Scorpion4ik [409]
3 years ago
11

Write the recursive, arithmetic and explicit formula for the nth term of the sequence:

Mathematics
1 answer:
Ivanshal [37]3 years ago
5 0

Recursive Formula

\begin{cases}a_1 = 8\\a_n = a_{n-1}-7\end{cases}

The top row says the first term is 8

The bottom row says that to get the nth term, we subtract 7 from the (n-1)th term. So basically we subtract 7 from each term to get the next term.

Note the subscripts tell us which term we're working with.

======================================================

Arithmetic Formula

We have a = 8 as the first term and d = -7 as the common difference.

a(n) = a + d(n-1)

a(n) = 8 + (-7)(n-1)

a(n) = 8 - 7n + 7

a(n) = -7n+15

The nth term arithmetic formula is a(n) = -7n+15

If you plug in n = 1, you should get a(n) = 8

If you plug in n = 2, you should get a(n) = 1

and so on.

=======================================================

Finding the 10th term

Plug in n = 10 to get

a(n) = -7n+15

a(10) = -7(10)+15

a(10) = -70+15

a(10) = -55

The 10th term is -55

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3 years ago
Given j(x) = x + 5, what is the value of j(12)?
yarga [219]

Answer:

j(12) = 17

General Formulas and Concepts:

<u>Pre-Algebra</u>

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<u>Algebra I</u>

  • Function Notation

Step-by-step explanation:

<u>Step 1: Define</u>

j(x) = x + 5

j(12) is x = 12

<u>Step 2: Evaluate</u>

  1. Substitute in <em>x</em>:                    j(12) = 12 + 5
  2. Add:                                     j(12) = 17
8 0
3 years ago
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Solve: 75(7-3)<br>with work shown​
anzhelika [568]

Answer:

300 is the solution

Step-by-step explanation:

use the distributive property

75 * 7 = 525

75 * -3 = -225

put em together

525 - 225 = 300

4 0
3 years ago
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Two positive integers have a sum of 21 and a product of 108. what are the integers
Vitek1552 [10]
X= first integer
21-x= second integer

(x)(21-x)= 108
multiply x by each term in parentheses

21x - x^2= 108
subtract 108 from both sides

21x - x^2 - 108= 0
-x^2 + 21x - 108= 0
multiply whole equation by -1 to eliminate the negative coefficient in the first term

(-1)(-x^2 + 21x - 108)= 0
x^2 - 21x + 108= 0
factor (remember we need #s to add to 21 and multiply to 108)

(x-12)(x-9)= 0

x-12= 0
x= 12

x-9= 0
x= 9

CHECK:
9 + 12= 21
9 * 12= 108

ANSWER: The two positive integers that add to 21 and multiply to 108 are 12 and 9.

Hope this helps! :)
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4 years ago
Solve the following 5x - 4=2x + 8
boyakko [2]
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+ 4          +4    Add 4 to both sides
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-2x       -2x     Subtract 2x from both sides
3x=12
/3     /3           Divide by 3 on both sides
x=4
6 0
3 years ago
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