Question 95 pts

alvin ran a 5-kilometer race.when he was halfway to the finish line, how many meters did he have left to run?

MrRissso [65]

Answer:

2.5km

step-by-step explanation:

half of 5

4 0

3 years ago

Ricardo throws a stone off a bridge into a river below. The stone's height (in meters above the water), xxx seconds after Ricard

Dominik [7]

For this case we have the following function:
w (x) = - 5 (x-8) (x + 4)
Rewriting we have:
w (x) = - 5 (x ^ 2 + 4x - 8x - 32)
w (x) = - 5x ^ 2 - 20x + 40x + 160
w (x) = - 5x ^ 2 + 20x + 160
Then, deriving we have:
w '(x) = - 10x + 20
We equal zero and clear x:
0 = -10x + 20
10x = 20
x = 20/10
x = 2 seconds
Substituting values:
w (2) = - 5 (2-8) (2 + 4)
w (2) = - 5 (-6) (6)
w (2) = 180 meters
Answer:
The maximum height that the stone will reach is:
w (2) = 180 meters

7 0

3 years ago

What is the best order-of-magnitude for 0.000006?

seraphim [82]

So,

Optimally, we want a number between 1 and 10 multiplied by a power of ten.

If we move the decimal place 6 places to the right, we get 6.

So we want $10^{-6}$ .

The correct option is C.

7 0

3 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

4 years ago

Can you please help me solve this I need it before 12/21/20 please help

Triss [41]

Answer:

I only know that the first box is and it is counting by twos

Step-by-step explanation:

because 5+2 =7 then 6+4=10 then 7+6=13 then 8+8=16 and lastly 9+10=19

7 0

3 years ago