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Maslowich
3 years ago
13

Write the slope-intercept form of the equation that passes through the point (4, -2) and is perpendicular to the line y =4x - 1

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
4 0

Answer:

<h2>         y = -¹/₄x - 1</h2>

Step-by-step explanation:

If two lines are perpendicular then the product of theirs slopes is equal -1:

y=m₁x+b₁   ⊥   y=m₂x+b₂   ⇒    m₁×m₂ = -1

y = 4x - 2   ⇒   m₁=4

4×m₂ = -1        ⇒    m₂ = -¹/₄

The line y=-¹/₄x+b passes through point (4, -2) so the equation:

-2 = -¹/₄×4 + b must be true

-2 = -1 + b

b = -1

Therefore equation in  slope-intercept form:

                                                y = -¹/₄x - 1

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Answer:

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Step-by-step explanation:

we know that

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so

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klasskru [66]

Answer:

Step-by-step explanation:

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6 0
2 years ago
Hi i just need help checking if my answer is correct I have been working on a implicit diffrention problem and I need some help.
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Answer:

  \dfrac{dy}{dx}=\dfrac{-4x^3y+2y^3+3x^2y^3}{x^4-6xy^2-3x^3y^2}

Step-by-step explanation:

Normally, when I do this, I differentiate each term first with respect to x then with respect to y. In this solution, I differentiated the entire expression with respect to x, then with respect to y. That makes separating the dx and dy coefficients much easier, so the solution almost falls into your lap.

  -x^4 y + 2 x y^3 + x^3 y^3=0 \qquad\text{given}\\\\(-4x^3y+2y^3+3x^2y^3)dx+(-x^4+6xy^2+3x^3y^2)dy=0\\\\(-4x^3y+2y^3+3x^2y^3)dx=(x^4-6xy^2-3x^3y^2)dy\\\\\boxed{\dfrac{dy}{dx}=\dfrac{-4x^3y+2y^3+3x^2y^3}{x^4-6xy^2-3x^3y^2}}

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