Find the unit rate. 21 3/4 meters in 2 1/2 hours

If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form

algol13

Answer:

$\frac{d^2y}{dx^2} = \frac{-4}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

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

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Basic Power Rule]: $-y'-6x^2=2yy'$
[Algebra] Isolate y' terms: $-6x^2=2yy'+y'$
[Algebra] Factor y': $-6x^2=y'(2y+1)$
[Algebra] Isolate y': $\frac{-6x^2}{(2y+1)}=y'$
[Algebra] Rewrite: $y' = \frac{-6x^2}{(2y+1)}$

Step 3: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}$
[Derivative] Back-Substitute y': $y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}$

Step 4: Find Slope at Given Point

[Algebra] Substitute in x and y: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Multiply: $y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}$
[Pre-Algebra] Divide: $y''(-1,-2) = \frac{-36+24 }{9}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-12}{9}$
[Pre-Algebra] Simplify: $y''(-1,-2) = \frac{-4}{3}$

6 0

2 years ago

What is the answe please help me

scZoUnD [109]

Answer:

Opcion C

Step-by-step explanation:

I think Maybe it could be

5 0

3 years ago

I only need 12 and 17, first good response i will give brainliest please i’m desperate

Mumz [18]

Answer:

$\textsf{12.} \quad y = 6x - 5$

$\textsf{17.} \quad y=-\dfrac{1}{4}x-2$

Step-by-step explanation:

<h3>Question 12</h3>

Find the slope of the line by substituting two points from the given table into the slope formula.

Define the points:

Let (x₁, y₁) = (2, 7)
Let (x₂, y₂) = (3, 13)

$\implies \textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{13-7}{3-2}=\dfrac{6}{1}=6$

Substitute the found slope and point (2, 7) into the point-slope formula to create an equation of the line:

$\implies y-y_1=m(x-x_1)$

$\implies y-7=6(x-2)$

$\implies y-7=6x-12$

$\implies y=6x-5$

<h3>Question 17</h3>

Given:

f(4) = 3
f(0) = -2

Therefore, two points on the line are:

(4, -3)
(0, -2)

The y-intercept is the y-value when x = 0.

Therefore, the y-intercept of the line is -2.

$\boxed{\begin{minipage}{6.3 cm}\underline{Slope-intercept form of a linear equation}\\\\$y=mx+b$\\\\where:\\ \phantom{ww}$\bullet$ $m$ is the slope. \\ \phantom{ww}$\bullet$ $b$ is the $y$-intercept.\\\end{minipage}}$