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3 years ago

The point (6, 5) lies on a circle centered at (3, –2). Find the radius of the circle.

1 answer:

3 0

The radius of circle is simply distance from center of circle to any point on the circle. since we have coordinates of both center and some point we can use formula for distance between 2 points in coordinate system to find radius.

R = √((6-3)^2 + (5-(-2))^2)
R = √(9 + 49)
R = √58

R = 7.6157 which round is 7.62

Answer is R = 7.62

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X = 0.5<br> Solve please

faltersainse [42]

Answer:

2x-1=0

Step-by-step explanation:

x-0.5=0

10x-5=0

2x-1=0

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3 years ago

One grade of tea costing $3.20 per pound is mixed with another grade costing $2 per pound to

wariber [46]

Answer:

12 of $3.2 grade is needed

Step-by-step explanation:

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3 years ago

Paula bought a jacket on sale for $6 less than half its original price. She paid $36 for the jacket. What was the original price

Fudgin [204]

Answer:

The correct answer is $30

Step-by-step explanation:

36-6=30

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4 years ago

Lucy is laying tile around the edge of a counter that is 6.4 feet by 2.7 feet. If each tile is 6 inches long, and Lucy buys 30 t

polet [3.4K]

Answer:

Tiles won't be enough

Step-by-step explanation:

The perimeter of the area :

2(length + Width)

Given dimension = 6.4 by 2.7 feets

Perimeter = 2(6.4 + 2.7) = 18.2 feets

1 feet = 12 inches

Perimeter = 18.2 * 12 = 218.4 inches

1 tile = 6 inches

30 tiles = 6* 30 = 180 inches

Since the perimeter of the area is greater Than the 180, then we can conclude that the tiles won't be enough

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3 years ago

Let F=(2x,2y,2x+2z)F=(2x,2y,2x+2z). Use Stokes' theorem to evaluate the integral of FF around the curve consisting of the straig

Brilliant_brown [7]

Stokes' theorem equates the line integral of $\vec F$ along the curve to the surface integral of the curl of $\vec F$ over any surface with the given curve as its boundary. The simplest such surface is the triangle with vertices (1,0,1), (0,1,0), and (0,0,1).

Parameterize this triangle (call it $T$ ) by

$\vec s(u,v)=(1-v)((1-u)(1,0,1)+u(0,1,0))+v(0,0,1)$

$\vec s(u,v)=((1-u)(1-v),u(1-v),1-u+uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $T$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,1-v,1-v)$

Divide this vector by its norm to get the unit normal vector. Note that this assumes a "positive" orientation, so that the boundary of $T$ is traversed in the counterclockwise direction when viewed from above.

Compute the curl of $\vec F$ :

$\vec F=(2x,2y,2x+2z)\implies\mathrm{curl}\vec F=(0,-2,0)$

Then by Stokes' theorem,

$\displaystyle\int_{\partial T}\vec F\cdot\mathrm d\vec r=\iint_T\mathrm{curl}\vec F\cdot\mathrm d\vec S$

where

$\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\,\mathrm dS$

$\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\mathrm d\vec S=\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

The integral thus reduces to

$\displaystyle\int_0^1\int_0^1(0,-2,0)\cdot(0,1-v,1-v)\,\mathrm du\,\mathrm dv=\int_0^12(v-1)\,\mathrm dv=\boxed{-1}$

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3 years ago