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What is longitude in spherical trigonometry?
It is a semi-circular arc joining two diametrically opposite points on the sphere. These may be the "North" and "South" poles.
Is Madara actually Tobi?
Yeah.....Madara did it so that the Hidden Leaf Village or some people won't not come after him --------------------------- Actually, no, swe found out in semi-recent manga chapters that Uchiha Madara was resurrected by Yakushi Kabuto. Tobi is nobody.
Does Pythagoras have a middle name?
No. For that matter, he didn't really have a last name. His name was Pythagoras. If anyone wanted to make sure that he didn't get confused with some other Pythagoras, they'd say something like "Pythagoras of Samos" (except, of course, that they'd say it in Greek: Î?υθαγόρας ὁ Σάμιος).The reason we have last names, middle names, etc. now is to avoid any confusion of this sort. Pointing up the need is the fact that there actually was another (semi-) famous Pythagoras living in Samos who was a sculptor and, to make matters worse, according to Pliny the Elder (see, there's another one we need to add a distinctive to; the modern system is looking smarter and smarter) he apparently looked a lot like the philospher/mathematician.
How can you use the Cosine Rule to prove Heron's Formula?
We stat with the law of cosines, which we can assume to be true: * c2 = a2 + b2 - 2ab*cos(C) Then rearrange it: * cos(C) = a2 + b2 - c2/2ab Use the identity sin(x)=SQRT(1-cos(x)) * sin(C) = SQRT( 1 - (a2 + b2 - c2/2ab)2) Use the operator A = 1/2ab*sin(C) where A is area. Also, set one equal to 4a2b2 and factor it out. * 2A/ab = SQRT(4a2b2 - (a2 + b2 - c2)2)/2ab ab's cancel, and the term inside the square root is the difference of two squares. * A = 1/4*SQRT((2ab - (a2 + b2 - c2))(2ab + (a2 + b2 - c2))) when the two groups are simplified, the can be factored in binomial squares. * A = 1/4*SQRT((c2 - (a - b)2)((a + b)2 - c2) Once again, we have differences of squares. * A = 1/4*SQRT((c - (a - b))(c + (a - b))((a + b) - c)((a + b + c)) Simplify. * A = 1/4*SQRT((c + b - a)(c + a - b)(a + b - c)(a + b +c)) Here comes the tricky part. We have four parts here. Three have two terms positive and one negative. Having a + b + c is like having the P. If we have a + b - c, that is like saying P - 2c, right? So to make it even easier, we can call s, the semi-perimeter, P/2. Then we can say a + b - c is 2s - 2c, or 2(s - c). We can apply that to all parts except the last one, which is just 2s. * A = 1/4*SQRT(2(s - a)*2(s - b)*2(s - c)*2s) The two's can multiply together to 16 and come out of the root, canceling with the 1/4. we are left with good old Heron's formula. * A = SQRT(s(s - a)(s - b)(s - c))
How many lines of symmetry does a semi circle?
a semi-circle only has one line of symmetry
How many lines of symmetry do semi circle have?
1
How many lines of symmetry does a cimmey circle has?
If "cimmey" is meant to be "semi" then the answer is 1. If not, I am sorry, I have no idea.
How many order of rotational symmetry does a semi circle have?
None.
How many NUMBER OF line of symmetry is semi circle?
Just the one and it is its radius when at right angles to the semi-circle's centre
How many axes of symmetry does a semi circle have?
one
How many parallel sides does a semi circle have?
a semi - circle has zero parallel lines
Where is the line of symmetry on a semi circle?
about its radius
How many parallel lines does a semi circle have?
None.
How many rotational symmetries does a semi circle have?
None, however the semicircle has one folding axis of symmetry perpendicular to the midpoint of the straight side
Does a semi circle have rotational symmetry?
If the central point of the straight line is placed exactly on the middle, and such central point has an axis, it will have a rotational symmetry.
How do you define a semi circle?
The semi circle is a shape which has 1 line of symmetry 1straight side 0 equal length sides 0 parallel sides 2 vertices 0 diagnals
