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3 years ago

Kalida runs 3 laps in 9 minuets and sonya runs 2 laps in 7 minuets who can run faster?

Mathematics

2 answers:

Vika [28.1K]3 years ago

8 0

Answer:

Kalida

Step-by-step explanation:

Kalida : 9 minutes / 3 laps = 3 minutes/ lap

Sonya : 7 minutes / 2 laps = 3.5 minutes/ lap

garri49 [273]3 years ago

5 0

Kalida because 9/3 =3/1 = she ran one lap in three minutes

Sonya is slower because she ran one lap in 4.5 minutes

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Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

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Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

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3 years ago

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iragen [17]

Answer: her monthly payments would be $267

Step-by-step explanation:

We would apply the periodic interest rate formula which is expressed as

P = a/[{(1+r)^n]-1}/{r(1+r)^n}]

Where

P represents the monthly payments.

a represents the amount of the loan

r represents the annual rate.

n represents number of monthly payments. Therefore

a = $12000

r = 0.12/12 = 0.01

n = 12 × 5 = 60

Therefore,

P = 12000/[{(1+0.01)^60]-1}/{0.01(1+0.01)^60}]

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3 years ago

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