All categories

3 years ago

Watz is one other way to name 1/3

Mathematics

1 answer:

Vlada [557]3 years ago

7 0

Is this in regards to probability?
The ones I can think of are one third, 1 to 3, and 1:3

You might be interested in

a triangle has one angle that measures 32 degrees and one angle that measures 102degrees. the measure of the triangle's thrid an

shutvik [7]

Step-by-step explanation:

all angles must add to 180

32+102= 134

180-134= 46

correct me if im wrong:)

3 0

3 years ago

Yooo can somebody help ill give you 10 or 22 points pls I need help!!!

dezoksy [38]

Answer:

The answer is 1475 have a nice day! <3

5 0

2 years ago

Read 2 more answers

Can I get some help plz

saul85 [17]

The line y=-x-3 begins in Q II and ends in Q IV. Since you have the "equal to or greater than" operator, the shaded area is ABOVE this line, and the line itself is dark (shaded) because of the "equal to."

4 0

3 years ago

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

sattari [20]

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

4 0

3 years ago

Given that ∠2 ≅ ∠10. Which statement is true?

Stolb23 [73]

Answer:

C) allb; by the converse alternate interior angels theorem

8 0

3 years ago