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3 years ago

Simplify to create an equivalent expression. 8(10 - 6q) + 3(-7q- 2) Choose 1 answer:

1 answer:

eduard3 years ago

7 0

Answer: -69q+74

Step by step:
1. Use distributive property.

8⋅10+8(−6q) + 3(−7q − 2)

Multiply 8 • 10 then -6 • 8

80 - 48q + 3(-7q - 2)

2. Use distributive property again.

80 - 48q + 3(-7q - 2)

Multiply -7 • 3 then 3 • -2

80 - 48q -21q - 6

3. Simplify

Subtract 6 from 80 then subtract 21q from -48q.

-69q + 74

8(10 - 6q) + 3(-7q - 2) = -69q + 74
———
I hope this answers your question. :)

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Use the limit comparison test to determine whether ∑n=19∞an=∑n=19∞8n3−2n2+196+3n4 converges or diverges.

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Answer:

Diverges

General Formulas and Concepts:

Algebra I

Exponential Rule [Dividing]: $\displaystyle \frac{b^m}{b^n} = b^{m - n}$

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Series Convergence Tests

P-Series: $\displaystyle \sum^{\infty}_{n = 1} \frac{1}{n^p}$
Direct Comparison Test (DCT)
Limit Comparison Test (LCT): $\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \sum^{\infty}_{n = 19} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}$

Step 2: Apply DCT

Define Comparison: $\displaystyle \displaystyle \sum^{\infty}_{n = 19} \frac{n^3}{n^4}$
[Comparison Sum] Simplify: $\displaystyle \displaystyle \sum^{\infty}_{n = 19} \frac{1}{n}$
[Comparison Sum] Determine convergence: $\displaystyle \displaystyle \sum^{\infty}_{n = 19} \frac{1}{n} = \infty , \ \text{div by P-Series}$
Set up inequality comparison: $\displaystyle\frac{8n^3 - 2n^2 + 19}{6 + 3n^4} \geq \frac{1}{n}$
[Inequality Comparison] Rewrite: $\displaystyle n(8n^3 - 2n^2 + 19) \geq 6 + 3n^4$
[Inequality Comparison] Simplify: $\displaystyle 8n^4 - 2n^3 + 19n \geq 6 + 3n^4 \ \checkmark \text{true}$

∴ the sum $\displaystyle \sum^{\infty}_{n = 19} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}$ is divergent by DCT.

Step 3: Apply LCT

Define: $\displaystyle a_n = \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}, \ b_n = \frac{1}{n}$
Substitute in variables [LCT]: $\displaystyle \lim_{n \to \infty} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4} \cdot n$
Simplify: $\displaystyle \lim_{n \to \infty} \frac{8n^4 - 2n^3 + 19n}{6 + 3n^4}$
[Limit] Evaluate [Coefficient Power Rule]: $\displaystyle \lim_{n \to \infty} \frac{8n^4 - 2n^3 + 19n}{6 + 3n^4} = \frac{8}{3}$

∴ Because $\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} \neq 0$ and the sum $\displaystyle \sum^{\infty}_{n = 19} a_n$ diverges by DCT, $\displaystyle \sum^{\infty}_{n = 19} \frac{8n^3 - 2n^2 + 19}{6 + 3n^4}$ also diverges by LCT.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Convergence Tests (BC Only)

Book: College Calculus 10e

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