Ask question

Ask question

All categories

Sergeeva-Olga [200]

3 years ago

5

The table shows the completion times of four runners in a race. Which list shows the runners in order by their completion times

from fastest to slowest
A. Patty, Joe, Steve, Ellen

B. Ellen, Joe, Steve, Patty

C. Ellen, Steve, Joe, Patty

D. Patty, Steve, Joe, Ellen

Please hurry its urgent

1 answer:

KengaRu [80]3 years ago

8 0

Answer:B

Step-by-step explanation:

Ellen, Joe, Steve, Patty

You might be interested in

Jasmine is getting a new playground in her backyard she graphed them playground area on the coordinate grid shown blow Each squa

Serhud [2]

The answer would be 20 /)

7 0

2 years ago

Read 2 more answers

Can someone help please?

ratelena [41]

Answer: It is TRUE. A point in a linear equation can be the solution to said equation.

4 0

3 years ago

Point C has the coordinates (–1, 4) and point D has the coordinates (2, 0). What is the distance between points C and D?

Bad White [126]

Answer:

5

Step-by-step explanation:

Use the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

insert the numbers into the formula $d=\sqrt{(2+1)^2+(0-4)^2}$

2+1=3 and 0-4=-4

$3^2$ =9 and $-4^{2}$ = 1

9+16=25

the square root of 25 is 5

6 0

3 years ago

Read 2 more answers

Write each fraction as a percent 2 __ 10

fgiga [73]

2_10 i'm guessing means 2/10, so in percentage, it would be 20%

5 0

3 years ago

The third-degree Taylor polynomial about x = 0 of In(1 - x) is

gizmo_the_mogwai [7]

Answer:

$\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Algebra I</u>

Functions
Function Notation

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

MacLaurin/Taylor Polynomials

Approximating Transcendental and Elementary functions
MacLaurin Polynomial: $\displaystyle P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... + \frac{f^{(n)}(0)}{n!}x^n$
Taylor Polynomial: $\displaystyle P_n(x) = \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n$

Step-by-step explanation:

*Note: I will not be showing the work for derivatives as it is relatively straightforward. If you request for me to show that portion, please leave a comment so I can add it. I will also not show work for elementary calculations.

<u />

<u>Step 1: Define</u>

<em>Identify</em>

f(x) = ln(1 - x)

Center: x = 0

<em>n</em> = 3

<u>Step 2: Differentiate</u>

[Function] 1st Derivative: $\displaystyle f'(x) = \frac{1}{x - 1}$
[Function] 2nd Derivative: $\displaystyle f''(x) = \frac{-1}{(x - 1)^2}$
[Function] 3rd Derivative: $\displaystyle f'''(x) = \frac{2}{(x - 1)^3}$

<u>Step 3: Evaluate Functions</u>

Substitute in center <em>x</em> [Function]: $\displaystyle f(0) = ln(1 - 0)$
Simplify: $\displaystyle f(0) = 0$
Substitute in center <em>x</em> [1st Derivative]: $\displaystyle f'(0) = \frac{1}{0 - 1}$
Simplify: $\displaystyle f'(0) = -1$
Substitute in center <em>x</em> [2nd Derivative]: $\displaystyle f''(0) = \frac{-1}{(0 - 1)^2}$
Simplify: $\displaystyle f''(0) = -1$
Substitute in center <em>x</em> [3rd Derivative]: $\displaystyle f'''(0) = \frac{2}{(0 - 1)^3}$
Simplify: $\displaystyle f'''(0) = -2$

<u>Step 4: Write Taylor Polynomial</u>

Substitute in derivative function values [MacLaurin Polynomial]: $\displaystyle P_3(x) = \frac{0}{0!} + \frac{-1}{1!}x + \frac{-1}{2!}x^2 + \frac{-2}{3!}x^3$
Simplify: $\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

Topic: AP Calculus BC (Calculus I/II)

Unit: Taylor Polynomials and Approximations

Book: College Calculus 10e

5 0

3 years ago