first three terms of a sequence are given. Round to the nearest thousand (if necessary) 152,149,146... Find the 30th term

1 answer:

4 0

Answer:

Step-by-step explanation:

use the following equation

aₙ=a₁+(n-1)d

Where d is the common difference

n is the term

a₁ is the first term

the common difference is -3 (149-152)

this means we have

152+(30-1)*-3

Compute this and get

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Algebra II

Step-by-step explanation:

Step 1: Define

Point A(2, 125)

Point B(98, 15)

Step 2: Identify

A(2, 125) → x₁ = 2, y₁ = 125

B(98, 15) → x₂ = 98, y₂ = 15

Step 3: Find distance d

Substitute in coordinates [Distance Formula]: $\displaystyle d = \sqrt{(98-2)^2+(15-125)^2}$
[√Radical] (Parenthesis) Subtract: $\displaystyle d = \sqrt{(96)^2+(-110)^2}$
[√Radical] Evaluate exponents: $\displaystyle d = \sqrt{9216+12100}$
[√Radical] Add: $\displaystyle d = \sqrt{21316}$
[√Radical] Evaluate: $\displaystyle d = 146$