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3 years ago

Helpppppp mathhhhhhhhhh

Mathematics

1 answer:

kogti [31]3 years ago

5 0

It would be HL theorem because the triangles are at 90 deg with a right angle

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Can someone help me???

mote1985 [20]

30 for the first and 83.5 for the second

7 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

The guitar that you always wanted is on sale in the music store. It costs 2/3 the original price. The guitar now costs $1996.67,

AysviL [449]

Answer:

3,327.9

Step-by-step explanation:

now: 1996.67

original: 3,327.9

1996.67/.6 = 3,327.9

4 0

3 years ago

How do i write this problem

ratelena [41]

3-(s+4)
less than- subtract
sum- addition

since it said 3 less than the sum of a number and 4 that means you would have to add s and 4

5 0

2 years ago

Read 2 more answers

a solid consists of cylinder surmounted by a right circular cone. The height of the cone is 2h. If the volume of the solid is 3

Mademuasel [1]

Answer:

height of cylinder = 4/3 h

Step-by-step explanation:

The solid has a cylinder surmounted with a cone .Therefore, the volume of the solid is the sum of the cone and the cylinder.

volume of the solid = volume of cylinder + volume of cone

volume of the solid = πr²h + 1/3πr²h

let

height of the cylinder = H

recall

the height of the cone = 2h

volume of the solid = πr²h + 1/3πr²h

3(1/3πr²2h) = πr²H + 1/3πr²2h

2πr²h = πr²H + 2/3 πr²h

πr²(2h) = πr²(H + 2/3 h)

divide both sides by πr²

2h = H + 2/3 h

2h - 2/3h = H

H = 6h - 2h/3

H = 4/3 h

height of cylinder = 4/3 h

8 0

2 years ago