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4 years ago

If sin Ѳ = cos 40, find the value Ѳ

Mathematics

2 answers:

Jet001 [13]4 years ago

7 0

Answer:

Θ = 50°

Step-by-step explanation:

Using the cofunction identity

cosx = sin(90 - x)

Given

cos40° = sin(90 - 40)° = sin50° ⇒ Θ = 50°

Andrej [43]4 years ago

7 0

Answer:

Ѳ = 50

Step-by-step explanation:

Sin Ѳ = cos 40

$Ѳ = \sin ^{ - 1} ( \cos(40) )$

Simplify

$Ѳ = sin^{ - 1} (0.7660444431) \\Ѳ = 50$

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Answer:

2 5/12

Step-by-step explanation:

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3 years ago

The Empire State Building weights about 7.3 x 10 ^8

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Answer:

730000000 lbs?

Step-by-step explanation:

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3 years ago

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. A. 139 B. 147° C. 155 D. 157 PLEASE HELP ASAAAPP!

enyata [817]

Answer:

$\huge\boxed{157 \ \ \text{degrees}}$

Step-by-step explanation:

In order to find the value of ∠EFB here, we have to note our angle relationships.

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Hope this helped!

5 0

3 years ago

A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R (> r) from th

jeyben [28]

Consider a circle with radius $r$ centered at some point $(R+r,0)$ on the $x$ -axis. This circle has equation

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$\displaystyle2\pi\int_R^{R+2r}2xy\,\mathrm dx$

where $2y=\sqrt{r^2-(x-(R+r))^2}-(-\sqrt{r^2-(x-(R+r))^2})=2\sqrt{r^2-(x-(R+r))^2}$ .

So the volume is

$\displaystyle4\pi\int_R^{R+2r}x\sqrt{r^2-(x-(R+r))^2}\,\mathrm dx$

Substitute

$x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt$

and the integral becomes

$\displaystyle4\pi r^2\int_{-\pi/2}^{\pi/2}(R+r+r\sin t)\cos^2t\,\mathrm dt$

Notice that $\sin t\cos^2t$ is an odd function, so the integral over $\left[-\frac\pi2,\frac\pi2\right]$ is 0. This leaves us with

$\displaystyle4\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}\cos^2t\,\mathrm dt$

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$\cos^2t=\dfrac{1+\cos(2t)}2$

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$\displaystyle2\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}(1+\cos(2t))\,\mathrm dt=\boxed{2\pi^2r^2(R+r)}$

6 0

3 years ago

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Wittaler [7]

Answer:

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Step-by-step explanation:

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3 years ago

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