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2 years ago

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in a petting zoo 1/4 of the animals are goats. 2/3 of the goats are white. what fraction of the petting zoo is white goats?in a

petting zoo 1/4 of the animals are goats. 2/3 of the goats are white. what fraction of the petting zoo is white goats?

1 answer:

Margarita [4]2 years ago

4 0

Im not exactly sure but it seems you have to add then divide ONCE again im not sure....

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Provide an example of a function that does not have an inverse function. How do you prove that it is not a function?

SOVA2 [1]

Answer:

f(x) = x^2 is a classic example

f( 3 ) = 9 and

f( -3 ) = 9, but

f^(-1)[9] is undefined because the result can be 3 and -3

6 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

A washing machine is filling up with water at a constant rate. The rate is 3/4 of a gallon in 1/5of a minute. Don calculated the

mel-nik [20]

He found out 1/5 of 3/4 instead of multiplying by 5

4 0

2 years ago

Read 2 more answers

Write the number 6\times 10^{-3}6×10 {−3}<br> in standard form.

solmaris [256]

Answer:

6 · 10⁻³

Step-by-Step Explanation:

The number 6 time 10 to the power -3 is 6 · 10⁻³ in standard form.

5 0

1 year ago

A swimming pool holds 540,000 liters of water. The pool has two drain pipes. When the pool is completely full, the first pipe al

tankabanditka [31]

Answer: 72 minutes

Step-by-step explanation:

Pipe A can empty the water in 180 minutes = (180/60) =3 hours

Pipe A does 1/ 3 of the job in 1 hour.

Pipe B can empty the water in 120 minutes = (120/60) = 2 hours.

Pipe B does 1/ 2 of thejob in 1 hour.

Together they will do:

1/ 3 + 1/ 2 = 5/6

Together, both pipes will do 5/ 6 of the job in one hour.

To get the total time to completely drain the water will be:

= 60 ÷ 5/6

= 60 × 6/5

= 12 × 6

= 72 minutes

It takes 72 minutes to empty the pool.

3 0

2 years ago