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3 years ago

Helppppppppp meeeeee

Mathematics

1 answer:

Margaret [11]3 years ago

4 0

Answer:

y=-1

Step-by-step explanation:

y=-1 just shows the y value which means its just a horizontal line which is parallel to the x axis.

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dem82 [27]

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3 0

3 years ago

to make his special drink, jerome uses 8 cups of water and 3 cups of drink mix. what is the ratio of water to drink mix?

Alisiya [41]

Answer:

8:3

Step-by-step explanation:

read the question carefully

3 0

3 years ago

Read 2 more answers

Ok look person that keeps answering Im sorry hut I gues u can tell why Im literally fai-ling math, question is: add the polynomi

Naddik [55]

Answer:

$21a^4+4a^2+2a$

Step-by-step explanation:

i think this is right..lol

hope it helps :)

have a nice day!!

7 0

3 years ago

I need help please <br>

Triss [41]

She paid $3.28 per gallon.

24.60 divided by 7.5 equals 3.28.

8 0

2 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

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

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

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,

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

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

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

6 0

3 years ago