Randy began completing the table below to represent a particular linear function. Write an equation to represent the

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

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

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

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:

We are given the following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$

Use the Basic Power Rule:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')$

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]$

Simplifying it, we have:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]$

We can rewrite the 2nd derivative using exponential rules:

$\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}$

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}$

When we evaluate this using order of operations, we should obtain our answer:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

5 0

3 years ago

Which expression is equivalent to the expression below? StartFraction 6 c squared 3 c Over negative 4 c 2 EndFraction divided by

Fudgin [204]

The given equation $(6c^2+3c)/(-4c+2)\div (2c+1)/(4c-2)$ is equivalent to the expression -3c.

Given that, the expression can be written as.

$(6c^2+3c)/(-4c+2)\div (2c+1)/(4c-2)$

By simplifying the above equation,

$\dfrac{6c^2+3c}{-4c+2}\div \dfrac{2c+1}{4c-2}$

By taking out the common terms from the equation,

$\dfrac{3c(2c+1)}{2(-2c+1)}\div\dfrac{2c+1}{2(2c-1)}$

By simplifying the above equation by cancel out the common factors.

$\dfrac{3c}{-2c+1} \div \dfrac{1}{2c-1}$

Now, by taking (-1) common from (-2c+1) we get,

$\dfrac{3c}{-1(2c-1)} \div \dfrac{1}{2c-1}$

By simplifying the above equation, we get the expression,

$-3c$

So the given equation $(6c^2+3c)/(-4c+2)\div (2c+1)/(4c-2)$ is equivalent to the expression -3c.

For more details, follow the link given below.

brainly.com/question/1301963.

7 0

3 years ago

What is the answer to this problem 75=3(-6n-5)

devlian [24]

Answer:

n=-5

Step-by-step explanation:

first you have to distribute the 3 to the -6 and the 5. In return u will get -18n and 15. then you have to cancel like terms. then you have to add 18 to the 75 which gives you 93. then you divide -18 by 93 which gives you -5.166 repeating... so all you have to do is round. your final answer will be -5.

5 0

3 years ago

on a map where each unit represents 100 miles , two airports are located at p(1,17) and q(12,10) what is the distance to the nea

dlinn [17]

Answer:

849 miles

Step-by-step explanation:

d = [(12-1)^2+(10-17)]^(1/2)

d = 8.49*100 = 849 miles.

7 0

3 years ago

Dosha is creating a new dessert that has two layers shaped like cones. The inner cone is frozen ice cream and has a diameter of

kotykmax [81]

Given:
Inner cone: diameter = 12 cm ; height = 6 cm
Outer layer: diameter = 12 cm ; height = 15 cm

Volume of a cone = π r² h/3

Inner cone: V = 3.14 * (6cm)² * 6cm/3 = 3.14 * 36cm² * 2cm = 226.08 cm³
Outer layer: V = 3.14 * (6cm)² * 15cm/3 = 3.14 * 36cm² * 5cm = 565.20 cm³

Volume of Outer layer : 565.20 cm³
less: Volume of inner layer: 226.08 cm³
Volume of cream filling: 339.12 cm³

5 0

3 years ago