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chubhunter [2.5K]
1 year ago
9

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

Mathematics
1 answer:
gavmur [86]1 year ago
5 0

The area of the part of the plane 3x 2y z = 6 that lies in the first octant  is  mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is  mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)

The partial derivatives of a function are f x and f y.

\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\

&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}

In conclusion,  the area is

A=3 √4 units ^2

Read more about the plane

brainly.com/question/1962726

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Kipish [7]

Answer:

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Step-by-step explanation:

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "p2"   was replaced by   "p^2".  1 more similar replacement(s).

STEP

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Equation at the end of step 1

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Equation at the end of step

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Checking for a perfect cube

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Trying to factor by pulling out :

3.2      Factoring:  4p3+6p2-8p-7

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -8p-7

Group 2:  4p3+6p2

Pull out from each group separately :

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Group 2:   (2p+3) • (2p2)

3.3    Find roots (zeroes) of :       F(p) = 4p3+6p2-8p-7

Polynomial Roots Calculator is a set of methods aimed at finding values of  p  for which   F(p)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  p  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

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The factor(s) are:

of the Leading Coefficient :  1,2 ,4

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Let us test ....

  P    Q    P/Q    F(P/Q)     Divisor

     -1       1        -1.00        3.00    

     -1       2        -0.50        -2.00    

     -1       4        -0.25        -4.69    

     -7       1        -7.00       -1029.00    

     -7       2        -3.50        -77.00    

     -7       4        -1.75        3.94    

     1       1        1.00        -5.00    

     1       2        0.50        -9.00    

     1       4        0.25        -8.56    

     7       1        7.00        1603.00    

     7       2        3.50        210.00    

     7       4        1.75        18.81    

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