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Juli2301 [7.4K]
3 years ago
11

Apply the product rules to determine the signof each expression ​

Mathematics
1 answer:
Inessa05 [86]3 years ago
4 0

Answer:

- + - = +

+ + + = +

+ + - = -

- + + = -

Step-by-step explanation:

Thanks hopes it helps

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You have 1/6 of a chance. 

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( 69 points + brain cuz why not xd )
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Answer:c

Step-by-step explanation:

The area of parallelogram A is;

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(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla
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Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}

    =\sqrt{1+1}\\\\=\sqrt{2}

calculating unit vector:

\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}

the point Q is at a distance h from P(6,6) Here, h=0.1  

a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071

the value of Q= (5.92928 ,6.07071  )

In point b:

Calculating the directional derivative of f (x, y) = \sqrt{x+3y} at P in the direction of \vec{v}

f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\

            =\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557

\vec{v} = 1.97651557

In point C:

Computing the directional derivative using the partial derivatives of f.

f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot   \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}

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