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kow [346]
3 years ago
13

Please hurry I need to pass this class

Mathematics
1 answer:
Dvinal [7]3 years ago
4 0

Answer:

\displaystyle m=\frac{-3}{4}

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> </u>

<u>Algebra I</u>

  • Reading a Cartesian plane
  • Coordinates (x, y)
  • Slope Formula: \displaystyle m=\frac{y_2-y_1}{x_2-x_1}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Find points from graph.</em>

Point A(0, 8)

Point B(4, 5)

<u>Step 2: Find slope </u><em><u>m</u></em>

Simply plug in the 2 coordinates into the slope formula to find slope <em>m</em>

  1. Substitute in points [Slope Formula]:                                                            \displaystyle m=\frac{5-8}{4-0}
  2. [Fraction] Subtract:                                                                                         \displaystyle m=\frac{-3}{4}
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Need help asap please!!
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8 0
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Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.
fredd [130]

\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}

\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

so the ODE is indeed exact and there is a solution of the form F(x,y)=C. We have

\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)

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8+\ln9=C

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\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

8 0
3 years ago
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