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3 years ago

What is the circumference of a circle with a radius of 2.5 meters?

Mathematics

1 answer:

Hitman42 [59]3 years ago

8 0

Answer:

15.71m

Step-by-step explanation:

circumference of a circle = 2πr

where r = 2.5m

where π = 3.142

we have,

2*3.142*2.5 = 15.71m

therefore the circumference of a circle with a radius 2.5m = 15.71 m

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Given that 5W = 2P + 3R find the value of P when W = 4 and R = −4

Olenka [21]

Answer:

P = 16

Step-by-step explanation:

Given

5W = 2P + 3R ← substitute W = 4 and R = - 4 into the equation

5(4) = 2P + 3(- 4), that is

20 = 2P - 12 ( add 12 to both sides )

32 = 2P ( divide both sides by 2 )

16 = P

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3 years ago

For this graph, mark the statements that are true.

Goshia [24]

Answer:

Step-by-step explanation:

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2 years ago

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liubo4ka [24]

First, set 2x+7 equal to 3x-7 since they are the same angle due to the parallel lines.

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2 years ago

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Step-by-step explanation:

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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

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

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

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

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2 years ago