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1 year ago

(2x+3y+4z)(4x^2+9y^2+16z^2-6xy-12yz-8zx)

1 answer:

6 0

1.1 Factoring: 4x2+9y2+16z2-6xy-12yz-8xz

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: -6xy-12yz
Group 2: 16z2-8xz
Group 3: 4x2+9y2

Pull out from each group separately :

Group 1: (x+2z) • (-6y)
Group 2: (x-2z) • (-8z)
Group 3: (4x2+9y2) • (1)

Looking for common sub-expressions :

Group 1: (x+2z) • (-6y)
Group 3: (4x2+9y2) • (1)
Group 2: (x-2z) • (-8z)

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30 2/9x3= 50 5/8 plz, plz, Thanks

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Step-by-step explanation:

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2 years ago

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2 years ago

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

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2 years ago

8.3 = -0.6 x + 11.3 what is x

finlep [7]

Answer:

x=5

Step-by-step explanation:

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2 years ago