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3 years ago

The circle below is centered at the point (4, 3) and has a radius of length 5. What is its equation?

Mathematics

2 answers:

kvasek [131]3 years ago

7 0

Answer:

(x-4)2+(y-3)2=5^2

Leona [35]3 years ago

6 0

Answer:

(x -4)² + (y -3)² = 25

Step-by-step explanation:

The circle centered at (h, k) with radius r has equation

... (x -h)² +(y -k)² = r²

Putting the given numbers into this formula gives

... (x -4)² +(y -3)² = 5²

... (x -4)² + (y -3)² = 25

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A chef is going to use a mixture two different brands of Italian dressing the first spring and days 5% vinegar the second brain

Temka [501]

I guess the chef is making the mixture for the sheriff... Let x be the amount of dressing with 5% vinegar that is required, and y the amount of 15% vinegar dressing (both amounts in mL).

The sheriff wants 390 mL of the mixed dressing, so that

x + y = 390

x mL of the 5% dressing contains 0.05x mL of vinegar, while y mL of the 15% dressing contains 0.15y mL of vinegar. The resulting mixture should have a concentration of 9% vinegar, so that it contains 0.09 (390 mL) = 35.1 mL of vinegar. This means

0.05x + 0.15y = 35.1

Solve for x and y :

y = 390 - x

0.05x + 0.15 (390 - x) = 35.1

0.05x + 58.5 - 0.15x = 35.1

23.4 = 0.10x

x = 234

y = 156

7 0

2 years ago

What is the y-intercept and slope of 3x-5y=15

vlada-n [284]

Answer:

Slope= 3/5

y-intercept= -3

Step-by-step explanation:

3 0

3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$