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3 years ago

How long will it take you to ski a distance of 36 miles at a speed of 3 miles per 30 minutes

1 answer:

7 0

36 ÷ 3 = 12
12 x 30 = 360 (mins)

The answer is that it would take you 360 minutes to ski a distance of 36 miles.
or you could say: t would take you 6 hours to ski a distance of 36 miles.

You might be interested in

PLEASE HELP! What is the volume of the pyramid shown below?

mezya [45]

The formula for the volume of a pyramid is,

V=1/3Bh

Since the base of the pyramid is a square, the base area is 11^2

which equals 121 sq. units

So, substitute 121

for B

and 6 for h, in the formula.

V=1/3(121)(6) = 242 cu. units

Or alternatively 1/3(11^2)(6) = 242 cu. units

8 0

2 years ago

Read 2 more answers

The heights of the trees in a forest are normally distributed, with a mean of 25 meters and a standard deviation of 6 meters. Wh

Alla [95]

Answer:

2.3%

Step-by-step explanation:

The formula to convert x into z distribution is

$z=\frac{x-\mu}{\sigma}$

Where z is the test statistic

x is what we are looking for (in this case 37)

$\mu$ is the mean (in this case, it is 25)

$\sigma$ is the standard deviation (we have 6)

Plugging these into the formula, we get:

$z=\frac{x-\mu}{\sigma}\\z=\frac{37-25}{6}\\z=2$

Thus we can say we want $P(z\geq 2)$

Note: $P(z\geq a)=1-P(z\leq a)$

The table given is for any z where $P(z\leq a)$

Thus, now we have:

$P(z\geq 2)\\=1-P(z\leq 2)\\=1-0.9772\\=0.0228\\$

0.0228 into percentage is 0.0228 * 100 = 2.28%

Rounded, we get 2.3%

Third answer choice is right.

3 0

3 years ago

Find the volume of a regular hexahedron if one of the diagonals of its faces is 8 root 2 inches.

Naily [24]

The answer is 114sqrt{6} in³

A regular hexahedron is actually a cube.

Diagonal of a cube D is a hypotenuse of a right triangle which other two legs are face diagonal (f) and length of a side (a):
D² = f² + a²

Face diagonal is a hypotenuse of a right triangle which sides are a and a:
f² = a² + a² =2a²

D² = f² + a²
f² = 2a²

D² = 2a² + a² = 3a²
D = √3a² = √3 * √a² = √3 * a = a√3

Volume of a cube with side a is: V = a³
D = a√3
⇒ a = D/√3

V = a³ = (D/√3)³

We have:
D = 8√2 in

$V= (\frac{D}{ \sqrt{3} } )^{3} =( \frac{8 \sqrt{2} }{ \sqrt{3} } )^{3}=( \frac{8 \sqrt{2} * \sqrt{3} }{ \sqrt{3}* \sqrt{3}} )^{3} =( \frac{8 \sqrt{2*3} }{ \sqrt{3*3}} )^{3} =( \frac{8 \sqrt{6} }{ \sqrt{3^{2} }} )^{3} =( \frac{8 \sqrt{6} }{ 3} )^{3} = \\ \\ = \frac{ 8^{3} *( \sqrt{6} )^{3}}{3^{3}} = \frac{512* \sqrt{6 ^{2} }* \sqrt{6} }{27} = \frac{512*6* \sqrt{6} }{27} = 114sqrt{6}$

8 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

3 years ago

Increase £139 by 28% Give your answer rounded to 2 DP

Elis [28]

Answer:

£177.92

Step-by-step explanation:

To increase something by 28% would be to multiply the original value by 1.28:

So to increase £139 by 28%, we do:

£139 × 1.28 = £177.92

So this is our final answer

4 0

2 years ago