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Nana76 [90]
2 years ago
5

Find the binomial coefficient: 2012/2011

Mathematics
1 answer:
seropon [69]2 years ago
4 0
 ²⁰¹²C₂₀₁₁ = (2012)! / [(2011)! (2012-2011)!]

²⁰¹²C₂₀₁₁ = (2012)! / [(2011)! (1)!]

Simplify 2012! / (2011)! = 2012

²⁰¹²C₂₀₁₁ = (2012)! / (1)!  = 2012
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Neko [114]

check the picture below.


so the cone itself has a height of 8, and a radius of 2.5, and the spherical scoop, ready to melt in it, has a radius of 2.5.


namely, will the volume of the sphere be larger than the cone? in which case the melted ice-cream will overflow.


\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=2.5\\ h=8 \end{cases}\implies V=\cfrac{\pi (2.5)^2(8)}{3}\implies \implies \boxed{V=\cfrac{50\pi }{3}} \\\\\\ \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}\qquad r=2.5\implies V=\cfrac{4(\pi )(2.5)^3}{3}\implies \boxed{V=\cfrac{125\pi }{6}} \\\\\\ \textit{the sphere's volume is larger, so yes, it will overflow}

8 0
2 years ago
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Rus_ich [418]
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An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side. 
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4 0
2 years ago
If jl= 10x-2 JK = 5x-8 and KL = 7x-12 find KL
natta225 [31]
JK+KL=JL
(5x-8)+7x-12=10x-2
(add common terms together)
12x-20=10x-2
(get the term with x alone on one side of the equation)
2x=18
(divide by 2 to get x alone)
x=9

Find KL:
KL=7(9)-12
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8 0
2 years ago
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lorasvet [3.4K]

Answer:

The puck traveled 53.8 meters.

Step-by-step explanation:

Currently, the player is 55 meters away from the opponent's goal.

After the shot, the puck is 120 centimeters = 1.2 meters from the opponents goal.

So, this means that the puck traveled 55 - 1.2 = 53.8 meters.

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1 year ago
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Answer:

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Step-by-step explanation:

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