12x+20=12x+20 how many solutions

Mathematics

1 answer:

WITCHER [35]3 years ago

6 0

Answer:

The equation is true

Step-by-step explanation:

Both sides are equal and they are all real numbers

Any value of x makes the equation true.

Hope this helps :)

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A test for marijuana usage was tried on 200200 subjects who did not use marijuana. the test result was wrong 77 times.

Alexxandr [17]

400,246 ( 200,200-77=200,123 divide by .5= 400,246

5 0

4 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$