Uh yeah? It's maximum/minimum (2U course)^ Is this a legit real Q. ???
1/2B = x (so that's done)I might as well start with a question for people to attempt.
Wow I really love that question, very unique in someway. Is there any chance that I could save her for you? No terms and conditions appliedUh yeah? It's maximum/minimum (2U course)
r is just the radius. So the height of the triangle is just r+y, and use pythagoras to find y, and then the solution falls out straight away.1/2B = x (so that's done)
to find the vertical height:
rotate r to make it horizontal the ad r + y which is the formula of the circle
there for h = (r + sq. root r^2 - x^2)
if you get what i mean
no need to explain to me, i wrote the question lol1/2B = x (so that's done)
to find the vertical height:
rotate r to make it horizontal the ad r + y which is the formula of the circle
there for h = (r + sq. root r^2 - x^2)
if you get what i mean
You made this question? It's very nice imo. I might come up with one of my own soon.no need to explain to me, i wrote the question lol
just for people to have a go at
but yes your right so far (and note what realisenothing said)
Yeh i did, there is a second part that i will post as soon as i solve it.You made this question? It's very nice imo. I might come up with one of my own soon.
nope you're correct for part bI got x=sqrt(3)/2 r
is that right so far?
probably made a silly mistake somewhere
oh goodnope you're correct for part b