Help me on this one

Mathematics

2 answers:

Gala2k [10]4 years ago

7 0

Set it as a proportion in minutes. 1500ft/60min=x ft/1min. You cross multiply to get 1500=60x, then you divide each side by 60 to get x=25 feet per minute

gulaghasi [49]4 years ago

5 0

25 ft/min. Just divide 1500 by 6.

You might be interested in

seraphim [82]

A parallel line has the same gradient

First you will need to rearrange the equation 10x+2y=-2

Then once you do that please comment

If you don’t know how please comment

7 0

3 years ago

Read 2 more answers

Please help will mark brainliest

TEA [102]

we know that

total of sum of arc is 360

so, sum of both arc must be 360

so, we get

$90+x=360$

now, we can solve for x

$90+x-90=360-90$

$x=270$ ............Answer

6 0

3 years ago

Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k

ch4aika [34]

Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

the curl of $\vec v$ is

$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$