As the title says, Scratch could use the ^ operator. As in:
<() ^ ()>And no, there is no valid workaround involving the repeat block. Reason why:
<(number) ^ ((1)/(3))>However,
repeat((1)/(3)) doSomething endis not going to work for obvious reasons. Scratch will not run something 1/3 of a time.
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Support! This is something that can only be done in BYOB, and even then it's not that easy and it's useful for finding distances as well as other things.
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Support! And thank you for teaching me how to find the square root/cube root of a number. I've been trying to find how forever and that's ALL it takes?!
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ImagineIt wrote:
Support! And thank you for teaching me how to find the square root/cube root of a number. I've been trying to find how forever and that's ALL it takes?!
Yeah, sqrt(x) is x^(1/2) and cube root of x is x^(1/3) etc.
I think you cover it in algebra so you might not have learned it.
I think there's a way to do exponents with logarithms as a workaround, but little kids probably won't get that so a ()^() block might be nice.
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What is the workaround? Just asking because I'm doing a project involving algebra in byob and an exponent block would be useful.
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shadowmouse wrote:
What is the workaround? Just asking because I'm doing a project involving algebra in byob and an exponent block would be useful.
kayybee wrote:
Yeah, sqrt(x) is x^(1/2) and cube root of x is x^(1/3) etc.
I think you cover it in algebra so you might not have learned it.
I think there's a way to do exponents with logarithms as a workaround, but little kids probably won't get that so a ()^() block might be nice.
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Mokat wrote:
shadowmouse wrote:
What is the workaround? Just asking because I'm doing a project involving algebra in byob and an exponent block would be useful.
kayybee wrote:
Yeah, sqrt(x) is x^(1/2) and cube root of x is x^(1/3) etc.
I think you cover it in algebra so you might not have learned it.
I think there's a way to do exponents with logarithms as a workaround, but little kids probably won't get that so a ()^() block might be nice.
I think you're missing the point.
shadowmouse wrote:
What is the workaround? Just asking because I'm doing a project involving algebra in byob and an exponent block would be useful.
There.
Last edited by lalala3 (2013-02-12 23:01:33)
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since everyone seems to be all excited about it, I guess I should post it.
I actually don't know/forgot it so ill have to think about it for a while.
Okay, I think I got it. it only took 6 minutes. Probably would've been faster searching I but oh well.
I think this should work. (you can use 10^ and log, or e^ and ln)
e^[y*ln(x)]=x^y
=10^[y*log(x)]
You might have to round it cause it'll be a bit off (like a billionth or a trillionth or so)
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Thanks Kayybee, I think Scratch rounds once it gets to ceratin amount of decimal places anyway.
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