PLZ HELP DUE SOON

Mr. Evans, my next door neighbour is fifty years old. He is five times older than his son, Lucas. Lucas is four years older than

OLEGan [10]

Answer:

Lilly is 7 years old

7 0

2 years ago

Answer???????????????????

andrezito [222]

∠ABC is 180°. ∠ABC-∠ABD=68°. Since line BE ∠DBC, 68 divide by 2 is 34°. So ∠EBC is <span>34°.</span>

7 0

2 years ago

There are five marbles in a bag two are yello and three are green. You draw a marble, replace it, and then draw another. What is

Sphinxa [80]

Answer: 2/5

Step-by-step explanation:

Since there are 5 marbles and 3 aren't yellow, you have a 2/5 chance to choose the other two yellow marbles

Hope this helps:)

5 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

To shop a package, a company charges a one-time fee plus a fee based on the weight of the package. The table below shows the tot

fiasKO [112]

Answer:

C = $5 + $1.5(w)

Step-by-step explanation:

Given the following information :

Total shipping cost :

One time fee + fee based on package weight

Given the table :

Weight in pounds - - - - Total shipping cost($)

___4__________________11

___8__________________17

___12_________________23

___16_________________29

We can deduce from the table

For a package that weighs (w) 4 pounds

Total shipping cost = $11

Let one time fee = f

Fee based on weight = r

f + 4(r) = 11 - - - - - (1)

For a package that weighs (w) 8 pounds

Total shipping cost = $17

One time fee = f

Fee based on weight = r

f + 8r = 17 - - - - - (2)

From (1)

f = 11 - 4r - - - (3)

Substitute f = 11 - 4r in (2)

11 - 4r + 8r = 17

-4r + 8r = 17 - 11

4r = 6

r = 6/4

r = 1.5

Put r = 1.5 in (3)

f = 11 - 4(1.5)

f = 11 - 6

f = 5

Hence one time fee = $5

Charge based on weight = $1.5

Hence, Total shipping cost 'C' for a package weighing 'w' will be :

C = $5 + $1.5(w)

7 0

3 years ago