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2 years ago

I need help please and thank you

Mathematics

2 answers:

Y_Kistochka [10]2 years ago

5 0

the surface area is 72 sq cm

hooe this helped

sammy [17]2 years ago

4 0

The answer is 72. You need multiply L x W x H. Because it is a rectangular prism and had 2 sides you can multiply 3 x 6 x 2 which equals 36 and then multiply 36 x 2 = 72

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balandron [24]

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2 years ago

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Rewrite 1/100,000,000 as a power of 10

yarga [219]

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True or False: The FOIL method is an example of the distributive property.

vova2212 [387]

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Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

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3 years ago

Trigonometry please help! work need to be shown

padilas [110]

Answer:

x = 5.56

Step-by-step explanation:

Apply trigonometric function to find x

Reference angle = 46°

Hypotenuse = 8

Adjacent length = x

Apply CAH:

Cos 46 = Adj/Hyp

Cos 46 = x/8

8 × Cos 46 = x

5.55726696 = x

x = 5.56 (nearest hundredth)

7 0

2 years ago