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Lemur [1.5K]
3 years ago
14

Anyone know what the answer is?

Mathematics
1 answer:
Citrus2011 [14]3 years ago
3 0
It’s been a minute but I think it’s x=3
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For what real values of does the quadratic 12x^2+kx+27=0 have nonreal roots
bija089 [108]

Answer:

k<±36

Step-by-step explanation:

Δ<0 (no real roots)

b²-4ac<0

k²-4x12x27<0

k²-1296<0

k²<1296

k<±36

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How many servings can be made from 108 cups of punch when you mix 18 cups of pineapple juice, 36 cups of orange juice and 54 cup
Alinara [238K]

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14


Step-by-step explanation:


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 (&gt; 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
2 years ago
Can someone help with this one
natali 33 [55]
Since the base is 4x4, and the height is 2, you'll use the 2 as the height of the netting and the 4 as the length. so, it will be 2x4 which is 8ft^2
4 0
2 years ago
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