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ser-zykov [4K]
3 years ago
11

Write the following numbers in order of size start with the smallest. 0.32 0.4 0.35 0.309

Mathematics
1 answer:
fenix001 [56]3 years ago
6 0

Answer:

Hey!

0.309<0.32<0.35<0.4

hope this helps you

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Function 1 is defined by the equation p= --3/2r - 5. function 2 is defined by the following table. Which function has a greater
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0-5. Here’s your answer
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The polar coordinates of a point are 3π 4 and 7.00 m. What are its Cartesian coordinates (in m)? (x, y) = 6.06,−3.5 m
Keith_Richards [23]

Answer:

a) \left(x,y\right)=\left(4.95,-4.95\right)

b) r\angle\theta = 7\angle0.5236\,\text{radians}

Step-by-step explanation:

Polar coordinates are represented as: r\angle\theta, where 'r' is the length (or magnitude) of the line, and '\theta' is the angle measured from the positive x-axis.

in our case:

7\angle\dfrac{3\pi}{4}

to covert the polar to cartesian:

x = r\cos{\theta}

y = r\sin{\theta}

we can plug in our values:

x = 7\cos{\dfrac{3\pi}{4}} = -7\dfrac{\sqrt{2}}{2}

y = 7\sin{\dfrac{3\pi}{4}} = 7\dfrac{\sqrt{2}}{2}

the Cartesian coordinates are:

\left(x,y\right)=\left(-7\dfrac{\sqrt{2}}{2},7\dfrac{\sqrt{2}}{2}\right)

\left(x,y\right)=\left(4.95,-4.95\right)

(b) to convert (x,y) = (6.06,-3.5)

we'll use the pythagoras theorem to find 'r'

r^2 = x^2+y^2

r^2 = (6.06)^2+(-3.5)^2

r = \sqrt{48.97} \approx 7

the angle can be found by:

\tan{\theta} = \dfrac{y}{x}

\tan{\theta} = \dfrac{3.5}{6.06}

\theta = \arctan{left(\dfrac{3.5}{6.06}\right)}

\theta = 0.5236 \text{radians}

to convert radians to degrees:

\theta = 0.5236 \times \dfrac{180}{\pi} \approx 30^\circ

writing in polar coordinates:

r\angle\theta = 7\angle30^\circ\,\,\text{OR}\,\,7\angle0.5236\,\text{radians}

5 0
3 years ago
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irga5000 [103]
We want to find a scalar function f(x,y) such that \nabla f(x,y)=\mathbf f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j.
So we need to have
\dfrac{\partial f}{\partial x}=3+4xy^2
Integrating both sides with respect to x gives
f(x,y)=3x+2x^2y^2+g(y)
Differentiating with respect to y gives
\dfrac{\partial f}{\partial y}=4x^2y+\dfrac{\mathrm dg}{\mathrm dy}=4x^2y\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C
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f(x,y)=3x+2x^2y^2+C
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klasskru [66]

Answer:

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