Find the equation of the line with slope m= -2 that contains the point (-8,18)

(35 points) answer 5 and 7 only please

tatyana61 [14]

A2+b2=c2 and get your answer

3 0

3 years ago

Solve 3-(2x-5)<-4(x+2)

azamat

Answer:

Step-by-step explanation:

3-(2x-5)=-4(x+2)

We simplify the equation to the form, which is simple to understand

3-(2x-5)=-4(x+2)

Remove unnecessary parentheses

3-2x+5=-4*(x+2)

Reorder the terms in parentheses

3-2x+5=+(-4x-8)

Remove unnecessary parentheses

+3-2x+5=-4x-8

We move all terms containing x to the left and all other terms to the right.

-2x+4x=-8-3-5

We simplify left and right side of the equation.

+2x=-16

We divide both sides of the equation by 2 to get x.

x=-8

7 0

3 years ago

Read 2 more answers

Estimate each product 5/7×1/9=

Cerrena [4.2K]

Multiply it out to get 5/7*1/9=5/63. This is an exact value of the product.

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

According to a survey by the National Center for Health Statistics, the heights of adult men in the U.S. are normally distribute

fiasKO [112]

Answer:

0.0151 is the probability that adult men are at most 59 inches or at least 74 inches.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 68 inches

Standard Deviation, σ = 2.75 inches

We are given that the distribution of heights of adult men is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(height between 59 and 74)

$P(59 \leq x \leq 74) = P(\displaystyle\frac{59 - 68}{2.75} \leq z \leq \displaystyle\frac{74-68}{2.75}) = P(-3.2727 \leq z \leq 2.1818)\\\\= P(z \leq 2.1818) - P(z < -3.2727)\\= 0.9854 - 0.0005=98.49\%$

$P(x\leq59 \cup x\geq 74)=1 - P(59\leq x \leq 74) \\=1 - 0.9849\\=0.0151\\=1.51\%$

0.0151 is the probability that adult men are at most 59 inches or at least 74 inches.

8 0

3 years ago