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SVEN [57.7K]
3 years ago
6

Answer yes or no to determine whether 57 is divisible by each number:

Mathematics
2 answers:
Dvinal [7]3 years ago
4 0

Answer:

2 no 3 yes 4 no 5 no 6 no 9 no and 10 no

Step-by-step explanation:

it only divisible by 3 if you're talking about whole numbers. Hope this helps!

Archy [21]3 years ago
3 0

Answer:

no this is not divisible

Step-by-step explanation:

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Cristina is sending out thank you cards for birthday presents. She has pink (P), blue (B), and green (G) cards, and white (W) an
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Answer:

PW, BW,  GW, PY,  BY, GY

Step-by-step explanation:

We need to determine the sample space

pink(P), blue (B), and green (G) cards,   (W) and yellow (Y) envelopes

Each color card can match with each color envelope

Start with the white envelopes and each color card

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Step-by-step explanation:

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You randomly draw twice from this deck of cards
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3 years ago
Calculate all four second-order partial derivatives and confirm that the mixed partials are equal. f(x,y)= 2e^xy
nadya68 [22]
ANSWER TO QUESTION 1

The given function is

f(x,y)=2 {e}^{xy}

The partial derivative of f with respect to x means we are treating y as a constant. The first derivative is

f_{x} = 2y {e}^{xy}

and the second derivative with respect to x is,

f_{xx} = 2 {y}^{2} {e}^{xy}

ANSWER TO QUESTION 2

The given function is

f(x,y)=2 {e}^{xy}

The partial derivative of f with respect to y means we are treating x as a constant. The first derivative is

f_{y} = 2x{e}^{xy}

and the second derivative with respect to y is

f_{yy} = 2 {x}^{2} {e}^{xy}

ANSWER TO QUESTION 3

Our first mixed partial is

f_{xy}

We need to differentiate
f_{x} = 2y {e}^{xy}
again. But this time with respect to y.

Since this is a product of two functions of y, we apply the product rule of differentiation to obtain,

f_{xy} = 2y( {e}^{xy})' + ({e}^{xy})(2y)'

f_{xy} = 2xy {e}^{xy} + 2{e}^{xy}

ANSWER TO QUESTION 4

The second mixed partial is

f_{yx}

We need to differentiate
f_{y} = 2x{e}^{xy}

again. But this time with respect to x.

Since this is a product of two functions of x, we apply the product rule of differentiation to obtain,


f_{yx} = 2x({e}^{xy})' + ({e}^{xy})(2x)'

f_{yx} = 2xy {e}^{xy} + 2{e}^{xy}

Hence,

f_{xy} = 2xy {e}^{xy} + 2{e}^{xy} =f_{yx}
4 0
3 years ago
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