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svlad2 [7]
3 years ago
11

Determine which description of description.

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

Answer:

Step-by-step explanation:

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Six friends together with their respective partners all meet up for a meal. To  commemorate the occasion, they arrange for a ph
olasank [31]

Answer:

1440 different arrangements are possible.

Step-by-step explanation:

Number of arrangements of n elements:

The number of arrangements of n elements is given by:

A_{n} = n!

In this question:

6 couples, in which each can be positioned in 2 ways(him/her or her/him). So

T = 2A_{6} = 2*6! = 1440

1440 different arrangements are possible.

8 0
3 years ago
Find the hcf 8x(x - 2) -2(x - 2)​
charle [14.2K]

We can see both terms have a (x-2) in them.

We can also see that 8 and 2 are both even so there is a 2 in common.

This means that the gcd or hcf or whatever you call it is 2(x-2)

**Just mentioning most of the time we call it the GCD(greatest common divisor) or the GCF(greatest common factor) and almost never uses hcf :)

3 0
3 years ago
Read 2 more answers
Can someone pls help me ASAP :/
Paraphin [41]

Answer:

d

Step-by-step explanation:

...........

..........

correct ans is D

4 0
3 years ago
Read 2 more answers
you are a geologist assigned to estimate the density of the Earth. look at picture for the full question:
Mila [183]
So, neverminding the units.

\bf g=\cfrac{GM}{R^2}\implies R^2=\cfrac{GM}{g}\implies R=\sqrt{\frac{GM}{g}}\quad 
\begin{cases}
G=6.67\cdot 10^{-11}\\
M=5.97\cdot 10^{24}\\
g=9.81
\end{cases}
\\\\\\
R=\sqrt{\cfrac{{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})}}{9.81}}
\\\\\\
R=\sqrt{\cfrac{{6.67\cdot 5.97\cdot 10^{13}}}{9.81}}
\\\\\\
R\approx 6371116.97~m\\\\
-------------------------------\\\\
V=\cfrac{4\pi R^3}{3}\implies V=\cfrac{4\pi 40591131498470.95^3}{3}
\\\\\\
V\approx 1083266582306022184268.1~m^3
6 0
3 years ago
Match each scenario with the object that will have the greatest gravitational force in that scenario
yarga [219]
Could you type them out of provide a different image? This one is difficult to read but I will gladly help you
8 0
3 years ago
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