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katrin2010 [14]
3 years ago
12

7 - 3x = 3 ( 3- x ) - 2 Please help & show how you got to the answer

Mathematics
2 answers:
Leokris [45]3 years ago
8 0
Step 1: Simplify both sides of the equation.
7−3x=3(3−x)−2
7+−3x=(3)(3)+(3)(−x)+−2(Distribute)
7+−3x=9+−3x+−2
−3x+7=(−3x)+(9+−2)(Combine Like Terms)
−3x+7=−3x+7
−3x+7=−3x+7
Step 2: Add 3x to both sides.
−3x+7+3x=−3x+7+3x
7=7
Step 3: Subtract 7 from both sides.
7−7=7−7
0=0
So all real numbers
Lostsunrise [7]3 years ago
6 0
First
7−3x=3(3−x)−2
7+−3x=(3)(3)+(3)(−x)+−2(Distribute)
7+−3x=9+−3x+−2
−3x+7=(−3x)+(9+−2)(Combine Like Terms)
−3x+7=−3x+7
−3x+7=−3x+7
secound 
−3x+7+3x=−3x+7+3x
7=7
third 
7−7=7−7
0=0 


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Step-by-step explanation:

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Step-by-step explanation:

Your question isn't really clear but let me help.

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Find the directional derivative of the function at the given point in the direction of the vector v. G(r, s) = tan−1(rs), (1, 3)
alexandr1967 [171]

The <em>directional</em> derivative of f at the given point in the direction indicated is \frac{5}{2}.

<h3>How to calculate the directional derivative of a multivariate function</h3>

The <em>directional</em> derivative is represented by the following formula:

\nabla_{\vec v} f = \nabla f (r_{o}, s_{o})\cdot \vec v   (1)

Where:

  • \nabla f (r_{o}, s_{o}) - Gradient evaluated at the point (r_{o}, s_{o}).
  • \vec v - Directional vector.

The gradient of f is calculated below:

\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{\partial f}{\partial r}(r_{o},s_{o})  \\\frac{\partial f}{\partial s}(r_{o},s_{o}) \end{array}\right]   (2)

Where \frac{\partial f}{\partial r} and \frac{\partial f}{\partial s} are the <em>partial</em> derivatives with respect to r and s, respectively.

If we know that (r_{o}, s_{o}) = (1, 3), then the gradient is:

\nabla f(r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{s}{1+r^{2}\cdot s^{2}} \\\frac{r}{1+r^{2}\cdot s^{2}}\end{array}\right]

\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{1+1^{2}\cdot 3^{2}} \\\frac{1}{1+1^{2}\cdot 3^{2}} \end{array}\right]

\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right]

If we know that \vec v = 5\,\hat{i} + 10\,\hat{j}, then the directional derivative is:

\nabla_{\vec v} f = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right] \cdot \left[\begin{array}{cc}5\\10\end{array}\right]

\nabla _{\vec v} f (r_{o}, s_{o}) = \frac{5}{2}

The <em>directional</em> derivative of f at the given point in the direction indicated is \frac{5}{2}. \blacksquare

To learn more on directional derivative, we kindly invite to check this verified question: brainly.com/question/9964491

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