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2 years ago

Given the two graphs below, identify in two complete sentences one similarity and one difference between the graphs.

Mathematics

1 answer:

Advocard [28]2 years ago

7 0

Answer:

well one similarity is that they both cross over zero and one difference is that one is going verticl and one is going up and down

Step-by-step explanation:

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Factor the expression completely by first taking out a common factor 3x^2 + 27x +24

Mrac [35]

The answer to this is
3(x+1)(x+8)

4 0

2 years ago

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

3 years ago

What percent is: x of y?

Zigmanuir [339]

Do you have a picture or something?

3 0

3 years ago

Hard math plz help!!!!!!!! PLZ NEED IT QUICK

Alex787 [66]

Answer:

Hi bub. Thanks for the points lol.

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Both circle A and circle B have a central angle measuring 140°. The ratio of the radius of circle A to the radius of circle B is

beks73 [17]

Let
rA--------> radius of the circle A
rB-------> radius of the circle B
LA------> the length of the intercepted arc for circle A
LB------> the length of the intercepted arc for circle B

we have that
rA/rB=2/3--------> rB/rA=3/2
LA=(3/4)π

we know that
if Both circle A and circle B have a central angle , the ratio of the radius of circle A to the radius of circle B is equals to the ratio of the length of circle A to the length of circle B
rA/rB=LA/LB--------> LB=LA*rB/rA-----> [(3/4)π*3/2]----> 9/8π

the answer is
the length of the intercepted arc for circle B is 9/8π

3 0

3 years ago