What's the square root of 18

How can you write the equation for a linear function if you know only two ordered pairs for the function?

Vinvika [58]

1. m
2. One set of ordered pairs
3. b

To show why this is, I’m going to explain how to write the equation for a linear function with two random sets of ordered pairs - (1,0) and (5, 8).

First, find the slope. The formula for slope is m = (y2 - y1)/(x2-x1) where m is the slope and (x1, y1) and (x2, y2) are two sets of points.

This is why #1 is m. M is the letter used when finding slope.

To find m, I plug in the two sets of ordered pairs.

m = (8-0)/(5-1)
m = 8/4
m = 2

An equation for a line (linear function) is written in something called slope-intercept form. It looks like y = mx + b. m is the slope and b is the y-intercept (number y equals when x = 0). If m = 3 and b = 1, the equation would be y = 3x + 1.

Here, you have just solved for m and know it equals 2. Plug that value in for m.

y = 2x + b

To solve for b, plug one ordered pair in for x and y. I will use (1,0)

0 = 2(1) + b
0 = 2 + b
-2 = b

Now that you know b = -2, plug that in for b.
y = 2x - 2. Now you have the equation fo the line.

4 0

2 years ago

PLEASE HELP ME GUYS OR I WONT PASS this calculus!!!!

KonstantinChe [14]

Answer:

b. $\displaystyle \frac{1}{2}$

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
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle H(x) = \sqrt[3]{F(x)}$

Step 2: Differentiate

Rewrite function [Exponential Rule - Root Rewrite]: $\displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}$
Chain Rule: $\displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)]$
Basic Power Rule: $\displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)$
Simplify: $\displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}$
Substitute in function values: $\displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}$
Exponents: $\displaystyle H'(5) = \frac{6}{3(4)}$
Multiply: $\displaystyle H'(5) = \frac{6}{12}$
Simplify: $\displaystyle H'(5) = \frac{1}{2}$