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Harrizon [31]
3 years ago
8

For what values of y:

Mathematics
1 answer:
Whitepunk [10]3 years ago
7 0

Answer:

y > \frac{3}{17}

Step-by-step explanation:

To solve this problem, simply set the binomial, 5y - 1 as greater than \frac{3y - 1}{4}, and solve for the value of y as you would solve an equation.

Thus:

5y - 1 > \frac{3y - 1}{4}

Multiply both sides by 4

4(5y - 1) > \frac{3y - 1}{4}*4

4*5y - 4*1 > 3y - 1

20y - 4 > 3y - 1

Subtract 3y from both sides

20y - 4 - 3y > 3y - 1 - 3y

17y - 4 > -1

Add 4 to both sides

17y - 4 + 4 > -1 + 4

17y > 3

Divide both sides by 17

\frac{17y}{17} > \frac{3}{17}

y > \frac{3}{17}

Therefore, the value of y that will make the binomial 5y - 1 greater than \frac{3y - 1}{4} is y > \frac{3}{17}.

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Answer:

See explanation.

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

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}}

<u>Calculus</u>

Differentiation

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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)]

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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 <em>x</em> = 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 <em>x</em> = 2, simply substitute in <em>x</em> = 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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