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svetlana [45]
3 years ago
5

The given line passes through the points and (4, 1).

Mathematics
2 answers:
pashok25 [27]3 years ago
5 0

Answer:

y-3=-2(x+4)

Step-by-step explanation:

<u><em>The complete question is</em></u>

The given line passes through the points (−4, −3) and (4, 1).

What is the equation, in point-slope form, of the line that is perpendicular to the given line and passes through the point (−4, 3)?

step 1

Find the slope of the given line

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

we have

(−4, −3) and (4, 1)

substitute the values

m=\frac{1+3}{4+4}

m=\frac{4}{8}

m=\frac{1}{2}

step 2

Find the slope of the perpendicular line to the given line

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

so

m_1*m_2=-1

we have

m_1=\frac{1}{2} ---> slope of the given line

so

m_2=-2 ---> slope of the perpendicular line to the given line

step 2

Find the equation in point slope form

y-y1=m(x-x1)

we have

m=-2

point\ (-4,3)

substitute

y-3=-2(x+4) ----> equation in point slope form

trasher [3.6K]3 years ago
5 0

Answer:

A

Step-by-step explanation:

did the edge quiz

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Hello again! This is another Calculus question to be explained.
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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

  • 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:

  1. f(x) = cxⁿ
  2. 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 <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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