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2 years ago

25 x 170 x 6.350 please help

Mathematics

2 answers:

m_a_m_a [10]2 years ago

8 0

Answer:

26,987.5

Step-by-step explanation:

hope it's helpful to you ☺️

$\sf{}$

♛┈⛧┈┈•༶♛┈⛧┈┈•༶

✌️

mario62 [17]2 years ago

4 0

Answer:

The answer is

26987.5

Hope this helps you

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What is the coefficient in this expression? 5 minus 4.7 minus 2 x + StartFraction 5 over 8 EndFraction

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Answer:

2 is the coefficient

Step-by-step explanation:

2 is the coefficient bc a coefficient is the number next to a variable (such as x) and 2 is next to x and is the only one in the equation

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- 7v7

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2 years ago

Show that if the vector field F = Pi + Qj + Rk is conservative and P, Q, R have continuous first-order partial derivatives, then

olchik [2.2K]

Answer:

It is proved that $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}$

Step-by-step explanation:

Given vector field,

$F=P\uvec{i}+Q\uvec{j}+R\uvec{k}$

Where,

$P=f_x=\frac{\partial f}{\partial x}, Q=f_y=\frac{\partial f}{\partial y}, R=f_z=\frac{\partial f}{\partial z}$

To show,

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}$

Consider,

$\frac{\partial P}{\partial y}=\frac{\partial}{\partial y}(\frac{\partial f}{\partial x})=\frac{\partial^2 f}{\partial y\partial x}=\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial y})=\frac{\partial Q}{\partial x}$

$\frac{\partial P}{\partial z}=\frac{\partial}{\partial z}(\frac{\partial f}{\partial x})=\frac{\partial^2 f}{\partial z\partial x}=\frac{\partial^2 f}{\partial x\partial z}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial z})=\frac{\partial R}{\partial x}$

$\frac{\partial Q}{\partial z}=\frac{\partial}{\partial z}(\frac{\partial f}{\partial y})=\frac{\partial^2 f}{\partial z\partial y}=\frac{\partial^2 f}{\partial y\partial z}=\frac{\partial}{\partial y}(\frac{\partial f}{\partial z})=\frac{\partial R}{\partial y}$