What Is the Pattern in Raven's Matrix Problem?

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Homework Statement



I'm not sure if this is the right part of forum to seek help with this type of problem, please correct me if this is the case.

As seen in attachment, I cannot figure out a pattern that I'm confident with. Please help with the answer and explain the reasoning, thanks!

Homework Equations



these problems can be solve horizontally, vertically or diagonally, or combinations of all

The Attempt at a Solution



suppose the question is split into 2 half: bottom half and top half



bottom half

by count:
3 circles
2 squares
1 triangles
2 diamonds
no conclusion

by number of vertices:
3-4-4
4-1-1 (take circle as 1)
4-1-1
guessing it is circle


top half

by horizontal spinning
1 a cross, turn 45 degree, then turn 45 degree
2 black dot spin, then unable to explain the horizontal line, then dot again
3 same as 1

by combine
doesn't work

by subtract
no clue



anyway, i think choice 1 is the most likely answer, but I'm not 100 % confident with it. Please help

most unsure about bottom half
 

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I have no idea what you are asking. You never say what the question itself is and, although you say 'i think choice 1 is the most likely answer' but you do not mention any choices!
 
Top halves are point symmetric.
Items with the same top have different bottoms.
 
HallsofIvy
Re: need help with Raven's matric problem
I have no idea what you are asking. You never say what the question itself is and, although you say 'i think choice 1 is the most likely answer' but you do not mention any choices!

Please look at the attachment, thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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