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P5641
Tue 2022-08-09 16:05:44
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07e18353ba2e769b832334d4da96566005c87867351e9e1e6997706540adb943.jpg
63.1 KiB 1024x1024
bonus points for elegance
P5643
Tue 2022-08-09 17:01:08
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d6374a51e19ba4865e2c77040b12438940681bc48c7d99e7930f2e482e838532.png
3.71 KiB 483x480
Let's call A the area of the square and A/5=a the area of each rectangle.
O = orange, Y = yellow, G = green, P = pink, B = blue
O+Y+G+P = 4a, so the side of the square is 8. A=8²=64
This was surprisingly easy? See the attached image for the sides of each rectangle.
Referenced by:
P5645
P5645
Tue 2022-08-09 17:05:18
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P5643
Yep, that's probably the easiest way to do it.
P5699
Wed 2022-08-10 00:24:50
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reply
[tex:
\frac
{
k
^
2
}
{
5
}
= 2n
]
[tex:
k
^
2
\frac
{
4
}
{
5
}
= k
\int
_
0
^
n
d
\xi
]
[tex:
k
^
2
\frac
{
4
}
{
5
}
= kn
]
[tex:
\frac
{
k
^
2
}
{
5
}
= 2k
\frac
{
4
}
{
5
}
]
[tex:
\text{Wait what was the question} \\ ?
]
Referenced by:
P5700
P5706
P5700
Wed 2022-08-10 00:29:56
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reply
>>
P5699
And can we get align / amsmath ? The reward for one good thing, etc.
P5706
Wed 2022-08-10 02:59:57
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reply
I felt like
P5699
was over defined. The problem describes one equation
[tex:
\int
_
0
^
n
kd
\xi
=4
\cdot
{2n}
]
Hence the area is
[tex:
k
^
2
=8
^
2
]
.
Mod Controls:
x
Reason: