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Dennis_Churaev [7]
3 years ago
13

Find the ratio of the following 2kg 250g to 3kg 500g​

Mathematics
1 answer:
goblinko [34]3 years ago
4 0

Answer:

4.5kg :7kg

Step-by-step explanation:

2 kg 250g convert in kg is 2.25kg

and 3kg 500g convert in kg is 3.5kg

2.25/3.5

solve the divide

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Miracle sells 28 dozens of donuts on Wednesday. Each dozen cost $18. How much is one
Nutka1998 [239]

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$1.5

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4 0
3 years ago
For each of the following vector fields
olga nikolaevna [1]

(A)

\dfrac{\partial f}{\partial x}=-16x+2y

\implies f(x,y)=-8x^2+2xy+g(y)

\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y

\implies\dfrac{\mathrm dg}{\mathrm dy}=10y

\implies g(y)=5y^2+C

\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}

(B)

\dfrac{\partial f}{\partial x}=-8y

\implies f(x,y)=-8xy+g(y)

\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x

\implies \dfrac{\mathrm dg}{\mathrm dy}=x

But we assume g(y) is a function of y alone, so there is not potential function here.

(C)

\dfrac{\partial f}{\partial x}=-8\sin y

\implies f(x,y)=-8x\sin y+g(x,y)

\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y

\implies\dfrac{\mathrm dg}{\mathrm dy}=4y

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\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}

For (A) and (C), we have f(0,0)=0, which makes C=0 for both.

4 0
3 years ago
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