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weeeeeb [17]
3 years ago
15

Which graph represents the solution to the system of inequalities below?

Mathematics
1 answer:
Reika [66]3 years ago
3 0

Answer:

The correct option is;

B

Step-by-step explanation:

The given system of inequalities are;

5·x - 4·y > 4...(1)

x + y < 2...(2)

Representing both inequalities as a function of "y", gives;

For, 5·x - 4·y > 4...(1), we have;

-4·y > 4 - 5·x

y < 4/(-4) - 5·x/(-4)

∴ y < 5·x/4 - 1

For x + y < 2...(2), we have;

y < 2 - x

Therefore, y is less than the values given by the equation of the straight line equalities, and the feasible region is given by the common region under both dashed lines representing both inequalities as shown in the attached diagram created using Microsoft Excel

The correct option is therefore, B.

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Can anybody help plzz?? 65 points
Yakvenalex [24]

Answer:

\frac{dy}{dx} =\frac{-8}{x^2} +2

\frac{d^2y}{dx^2} =\frac{16}{x^3}

Stationary Points: See below.

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Properties

<u>Calculus</u>

Derivative Notation dy/dx

Derivative of a Constant equals 0.

Stationary Points are where the derivative is equal to 0.

  • 1st Derivative Test - Tells us if the function f(x) has relative max or mins. Critical Numbers occur when f'(x) = 0 or f'(x) = undef
  • 2nd Derivative Test - Tells us the function f(x)'s concavity behavior. Possible Points of Inflection/Points of Inflection occur when f"(x) = 0 or f"(x) = undef

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

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:

<u>Step 1: Define</u>

f(x)=\frac{8}{x} +2x

<u>Step 2: Find 1st Derivative (dy/dx)</u>

  1. Quotient Rule [Basic Power]:                    f'(x)=\frac{0(x)-1(8)}{x^2} +2x
  2. Simplify:                                                      f'(x)=\frac{-8}{x^2} +2x
  3. Basic Power Rule:                                     f'(x)=\frac{-8}{x^2} +1 \cdot 2x^{1-1}
  4. Simplify:                                                     f'(x)=\frac{-8}{x^2} +2

<u>Step 3: 1st Derivative Test</u>

  1. Set 1st Derivative equal to 0:                    0=\frac{-8}{x^2} +2
  2. Subtract 2 on both sides:                         -2=\frac{-8}{x^2}
  3. Multiply x² on both sides:                         -2x^2=-8
  4. Divide -2 on both sides:                           x^2=4
  5. Square root both sides:                            x= \pm 2

Our Critical Points (stationary points for rel max/min) are -2 and 2.

<u>Step 4: Find 2nd Derivative (d²y/dx²)</u>

  1. Define:                                                      f'(x)=\frac{-8}{x^2} +2
  2. Quotient Rule [Basic Power]:                  f''(x)=\frac{0(x^2)-2x(-8)}{(x^2)^2} +2
  3. Simplify:                                                    f''(x)=\frac{16}{x^3} +2
  4. Basic Power Rule:                                    f''(x)=\frac{16}{x^3}

<u>Step 5: 2nd Derivative Test</u>

  1. Set 2nd Derivative equal to 0:                    0=\frac{16}{x^3}
  2. Solve for <em>x</em>:                                                    x = 0

Our Possible Point of Inflection (stationary points for concavity) is 0.

<u>Step 6: Find coordinates</u>

<em>Plug in the C.N and P.P.I into f(x) to find coordinate points.</em>

x = -2

  1. Substitute:                    f(-2)=\frac{8}{-2} +2(-2)
  2. Divide/Multiply:            f(-2)=-4-4
  3. Subtract:                       f(-2)=-8

x = 2

  1. Substitute:                    f(2)=\frac{8}{2} +2(2)
  2. Divide/Multiply:            f(2)=4 +4
  3. Add:                              f(2)=8

x = 0

  1. Substitute:                    f(0)=\frac{8}{0} +2(0)
  2. Evaluate:                      f(0)=\text{unde} \text{fined}

<u>Step 7: Identify Behavior</u>

<em>See Attachment.</em>

Point (-2, -8) is a relative max because f'(x) changes signs from + to -.

Point (2, 8) is a relative min because f'(x) changes signs from - to +.

When x = 0, there is a concavity change because f"(x) changes signs from - to +.

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