Identify the slope of the line y=-4- 1/6x

What is estimate 4.23-2.51

ruslelena [56]

The answer is 1.72 and it would be 2 if u estimated

4 0

3 years ago

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What is the difference between 27,202-3,489

insens350 [35]

27202 - 3489 you can't do 2-9 so you cross out the 0 next to it and put it into a 10 and then turn your to into a 12 so 12 - 9 is 3 then since you have your 0 turned into 10 you do 10-8 you would you get 2 then you can't do 2-4 so we got the cross out your 7 next to it and turn that into a 6 then change 2 into 12 and then do 12 - 4 equals 8 then since you changed your 7
into a 6 6-3 would you get three and then you do 2-0 because there is no number there and you put down 23823 so 23823 is your answer

7 0

3 years ago

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Josefina pain rent for 5 months. She paid a total of $2045. how much did she pay per month?

djverab [1.8K]

Answer:

$409 per month

Step-by-step explanation:

If josefina paid $2045 over a time period of 5 months, then we have to divide how much she paid ($2045) by 5 to find out how much she paid for rent in 1 month.

We can use the equation: 2045 ÷ 5

When we solve that equation, we get 409 which means that josefina paid $409 per month for rent.

(sorry if this is confusing but I hope it helped)

8 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:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions
Function Notation

Calculus

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.

Step 1: Define

Identify

f(x) = ln(1 - x)

Center: x = 0

n = 3

Step 2: Differentiate

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

Step 3: Evaluate Functions

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

Step 4: Write Taylor Polynomial

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

I’ll click the heart please help

Pavlova-9 [17]

Answer:

The answer is just 4

Step-by-step explanation:

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8 0

3 years ago

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