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Ugo [173]
3 years ago
7

The solutions of the equation x^2-6x+4=0

Mathematics
1 answer:
lesya692 [45]3 years ago
6 0

Answer:

you have 2 options!!!

Step-by-step explanation:

hoped I helped!

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BRAINLIEST ANSWERS!One water molecule
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Um I think 4 i hope this helps
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Two cats, Fluffy and Fireball, met in the park. Fluffy's tail is 1/3​​, of a meter long. Fireball's tail is 1/4 of a meter long.
Lunna [17]
1/3 - 1/4 = 4/12 - 3/12 = 1/12....Fluffy's tail is 1/12 meters longer then Fireball's tail
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Write the decimal 40.99 in word form
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Forty and ninety nine hundreths

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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

(x-(R+r))^2+y^2=r^2

Revolve the region bounded by this circle across the y-axis to get a torus. Using the shell method, the volume of the resulting torus is

\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

Write

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

so the volume is

\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
What is the value of x in the equation 1.5(x + 4) – 3 = 4.5(x – 2)?<br> a 3<br> b 4<br> c 5<br> d 9
Pepsi [2]
The answer would be B) x = 4.

1.5x + 6 - 3 = 4.5x - 9

1.5x + 3 = 4.5x - 9

3 = 4.5x - 9 - 1.5x

3 = 3x - 9

3 + 9 = 3x

12 = 3x

12/3 = x

4 = x

x = 4.
7 0
3 years ago
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