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mihalych1998 [28]
3 years ago
15

Evaluate 6C3. 18 20 60 120

Mathematics
1 answer:
melomori [17]3 years ago
4 0

Answer:

Step-by-step explanation:

nC_{r}=\frac{n!}{r!(n-r)!}\\\\6C_{3}=\frac{6!}{3!(6-3)!}\\\\=\frac{6!}{3!(6-3)!}==\frac{6!}{3!3!}\\=\frac{6*5*4*3!}{3!3!}\\\\=\frac{6*5*4}{3!}\\\\=\frac{6*5*4}{3*2*1}\\\\ =5*4=20

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3 years ago
What is the force exerted on an object<br> with a mass of 20 kg and an<br> acceleration of 6 m/s2?
OverLord2011 [107]

Answer:

F = 120 N

Step-by-step explanation:

Given that,

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F=ma\\\\F=20\ kg\times 6\ m/s^2\\\\F=120\ N

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3 years ago
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stepan [7]

Before answering this question let us know what are net and total areas.

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Now let us look at first part of our question. Net area of our region will be equal to total area of region when the whole region lies above x axis because then our whole region is positive.

The net area of a region will differ from the total area of the region when some part of our region is above x-axis and some part of region is below x-axis. This is because we have to subtract area of region which is below x-axis from area above x-axis to get net area.

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