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3 years ago

When plotting points on the coordinate plane below, which point would lie on the x-axis?

Mathematics

2 answers:

Margarita [4]3 years ago

4 0

(6,0) would land on the x- axis

nordsb [41]3 years ago

3 0

If you plot the points you can easily find the answer. (X,Y) The x-axis is the horizontal line and the y-axis is the vertical line.

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PLEASE HELP MEEEEE!!!!!!!

OverLord2011 [107]

Answer:

Cheese: $4\frac{3}{8}$
Milk: $1\frac{2}{3}$
Pasta: $6\frac{1}{4}$

Step-by-step explanation:

Cheese:

$1\frac{3}{4} = \frac{7}{4}$
Set up a proportion: $\frac{\frac{7}{4} }{4} = \frac{x}{10}$
Cross multiply, then divide: 10 × 7/4 = 17.5, 17.5 ÷ 4 = 4.375
$4.375 = 4\frac{3}{8}$

Milk:

Set up a proportion: $\frac{\frac{2}{3} }{4} = \frac{x}{10}$
Cross multiply, then divide: 10 × 2/3 = $6\frac{2}{3}$ , $6\frac{2}{3}$ ÷ 4 = $1\frac{2}{3}$

Pasta:

$2\frac{1}{2} = \frac{5}{2}$
Set up a proportion: $\frac{\frac{5}{2} }{4} = \frac{x}{10}$
Cross multiply, then divide: 10 × 5/2 = 25, 25 ÷ 4 = 6.25
$6.25 = 6\frac{1}{4}$

I hope this helps!

4 0

3 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.

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