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3 years ago

15/4÷3/4 I need the ans

2 answers:

7 0

Use the stay, change, flip method.
keep the first fraction the same, change the division to multiplication and flip the last fraction. you may then have the opportunity to cross simplify.

Alex3 years ago

3 0

First flip the second fraction so you can multiply
15/4➗4/3
Then cross multiply
5/1

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2 years ago

Find the directional derivative of at the point (1, 3) in the direction toward the point (3, 1). g

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Complete Question:

Find the directional derivative of g(x,y) = $x^2y^5$ at the point (1, 3) in the direction toward the point (3, 1)

Answer:

Directional derivative at point (1,3), $D_ug(1,3) = \frac{162}{\sqrt{8} }$

Step-by-step explanation:

Get $g'_x$ and $g'_y$ at the point (1, 3)

g(x,y) = $x^2y^5$

$g'_x = 2xy^5\\g'_x|(1,3)= 2*1*3^5\\g'_x|(1,3) = 486$

$g'_y = 5x^2y^4\\g'_y|(1,3)= 5*1^2* 3^4\\g'_y|(1,3)= 405$

Let P = (1, 3) and Q = (3, 1)

Find the unit vector of PQ,

$u = \frac{\bar{PQ}}{|\bar{PQ}|} \\\bar{PQ} = (3-1, 1-3) = (2, -2)\\{|\bar{PQ}| = \sqrt{2^2 + (-2)^2}\\$

$|\bar{PQ}| = \sqrt{8}$

The unit vector is therefore:

$u = \frac{(2, -2)}{\sqrt{8} } \\u_1 = \frac{2}{\sqrt{8} } \\u_2 = \frac{-2}{\sqrt{8} }$

The directional derivative of g is given by the equation:

$D_ug(1,3) = g'_x(1,3)u_1 + g'_y(1,3)u_2\\D_ug(1,3) = (486*\frac{2}{\sqrt{8} } ) + (405*\frac{-2}{\sqrt{8} } )\\D_ug(1,3) = (\frac{972}{\sqrt{8} } ) + (\frac{-810}{\sqrt{8} } )\\D_ug(1,3) = \frac{162}{\sqrt{8} }$

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