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2 years ago

23) A shopkeeper buys 1 dozen of pens at Rs. 20 each and sells at Rs. 18 each. Find his loss and loss percent. 24) The marl

Mathematics

2 answers:

kramer2 years ago

6 0

Answer:

loss = 24, loss % = 10

Step-by-step explanation:

amount he spent = 12*20 = 240

amount he earnt = 12*18 =216

loss = 240-216 = 24

loss % = 24/240 x 100% = 10%

GaryK [48]2 years ago

6 0

Answer:

yes

Step-by-step explanation:

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2 years ago

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3 years ago

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

Anika [276]

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)

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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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2 years ago

23) A shopkeeper buys 1 dozen of pens at Rs. 20 each and sells at Rs. 18 each. Find his loss and loss percent. 24) The marl​

23) A shopkeeper buys 1 dozen of pens at Rs. 20 each and sells at Rs. 18 each. Find his loss and loss percent. 24) The marl