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2 years ago

What were the total earnings of all five of these movies in the given week?

Mathematics

2 answers:

Tanya [424]2 years ago

7 0

Answer:

The earnings of all five movies were $154,960,863.44.

Step-by-step explanation:

Thats the answer from edmentum

babunello [35]2 years ago

5 0

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Answer: 500$

Explanation:

I hope this helped!

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- Zack Slocum

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3 years ago

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Cuanto es r2-2r-7=0

vekshin1

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Step-by-step explanation:

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3 years ago

If the volume of a cube is 64 in3, how long is each side?

8_murik_8 [283]

Answer:

Step-by-step explanation:

4*4*4=16*4=64

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3 years ago

In triangle EFG, angle E equals 8x-16, angle F equals x+8, and angle G equals 2x-10. What is the value of x?

Stels [109]

Answer: The value of x= 18

Step-by-step explanation:

Given: In triangle EFG, angle E equals 8x-16, angle F equals x+8, and angle G equals 2x-10.

The sum of all angles of a triangle is 180°.

Therefore, in triangle EFG

∠E + ∠F +∠G= 180°

⇒ 8x-16 +x+8+ 2x-10 = 180

⇒ 11x-18= 180

⇒ 11x = 180+18

⇒11x=198

⇒ x= 18 [Divide both sides by 11]

Hence, the value of x= 18

6 0

2 years ago

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

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

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

7 0

3 years ago