Need help with this Algebra sheet

A line goes through the points (5, -1) and (7, -5). What is a possible equation to a new line that is perpendicular?

Leya [2.2K]

Ok, so you are given an equation in standard form, 5x+4y=2, and a point, (7, 5). You are being asked to write the equation for a line that is parallel to the equation in standard from, and that includes the point (7, 5).

First, let's start by finding the slope of your new line. We know that it needs to have the same slope as 5x+4y=2, because parallel lines have the same slopes. To do that, we need to put the equation into slope-intercept form (y=mx+b), which means we need to isolate the "y."

5x+4y=2

-5x -5x (subtract 5x from both sides to move it to the right side of your equation)

4y = 2 - 5x

/4 /4 /4 (divide all the terms by 4 to get "y" by itself)

y = (1/2) - (5/4)x ... I suggest leaving your slope as a fraction.

Now, we know that our slope, m , is going to be -(5/4).

Next, we are going to use our slope, -(5/4), and point, (7, 5), to find the b-value (y-intercept) of your new line. Let's plug in what we know:

y = 5

m = -(5/4)

x = 7

y=mx+b

5=(-5/4)(7) + b -> I plugged in what we knew for y, m, and x.

5 = -(35/4) + b -> I multiplied the numbers in the numerator (5 x 7) to get 35/4

20/4 = -(35/4) + b -> I converted 5 into a faction with a denominator of 4 by multiplying by (4/4)

+(35/4) +(35/4) -> I add (35/4) to both sides to isolate b

55/4 = b ... or b = 13.74

Answer:

m = -(5/4)

b = 13.75

7 0

2 years ago

Implicit differentiation Please help

Anvisha [2.4K]

Answer:

$y''(-1) =8$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

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

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

Step-by-step explanation:

Step 1: Define

-xy - 2y = -4

Rate of change of the tangent line at point (-1, 4)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Product Rule/Basic Power Rule]: $-y - xy' - 2y' = 0$
[Algebra] Isolate y' terms: $-xy' - 2y' = y$
[Algebra] Factor y': $y'(-x - 2) = y$
[Algebra] Isolate y': $y' = \frac{y}{-x-2}$
[Algebra] Rewrite: $y' = \frac{-y}{x+2}$

Step 3: Find y

Define equation: $-xy - 2y = -4$
Factor y: $y(-x - 2) = -4$
Isolate y: $y = \frac{-4}{-x-2}$
Simplify: $y = \frac{4}{x+2}$

Step 4: Rewrite 1st Derivative

[Algebra] Substitute in y: $y' = \frac{-\frac{4}{x+2} }{x+2}$
[Algebra] Simplify: $y' = \frac{-4}{(x+2)^2}$

Step 5: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{0(x+2)^2 - 8 \cdot 2(x + 2) \cdot 1}{[(x + 2)^2]^2}$
[Derivative] Simplify: $y'' = \frac{8}{(x+2)^3}$