x and c-x, then the diagram looks as follows: The perpendicular has divided the triangle into two right-angled triangles. Now for any right-angle triangle, according to Pythagorean Theorem, [pic] = [pic] + [pic] If Pythagorean is applied to the right-angled triangles in the above triangle, then in the case of left right-angle triangle in the above diagram, it would give us the equation [pic] = [pic] + [pic] where ‘a’ = hypotenuse and ‘h’ = height/perpendicular and ‘x’ = base. Re-writing it, the equation would become which we will call Eq. A [pic] = [pic] - [pic] ---------------------( Eq. A Similarly, for the right angle triangle on the right half to triangle ABC, [pic] = [pic] + [pic] where ‘b’ = hypotenuse, ‘h’ = height/perpendicular and ‘c-x’ = base.
Notice that the angle of incidence at the first surface is equal to the angle of refraction at the second surface. Prove algebraically (using Snell’s Law twice) why this is true. n1sin θ1=n2sin θ2 n2sin θ2=n1sin θ3 n1sin θ1=n1sin θ3 sin θ1=sin θ3 5. Do different colors refract at different angles? No, difference colors do not refract at different angles.
5. Tin's actual atomic radius is 140 pm. Use the math equation below to determine the percent error of your estimated value. (140-145) x100 = 3.5 140 1. The atomic radius you estimated from your graph is most likely not exactly the same as the true atomic radius of tin.
What element in Period 5 of the periodic table is a member of Group 14? (1 point) Answer: The element that is in Period 5 as well as Group 14 is tin or Sn. 4. The word interpolate means to use a given line graph to find unknown points between the plotted points of the graph. Use your line graph from Part II to interpolate, or estimate, atomic radius of Tin (Sn). (1 point) Answer: By using the given line to interpolate the atomic radius of tin, it can be found that its radius is around 149pm.
Week 5 Law of Sines and Cosines 1. Law of Sines: An equation to the length of all sides on a triangle and to the sines of its angles. 2. Law of Cosines: Related to the lengths of the sides of a plane triangle to the cosine of one of the angles. 3.
The circles located on the inside are called inscribed circles, inscribed circles should touch each side of the triangle at a single point. The circles on the out sides are called circumscribed circles, the circumscribed circle touches each vertex of the triangle. There are also things called the incenter and the circumcenter. The incenter is the center of the inscribed circle and the circumcenter is the center of a circumscribed circle,the relation between a circle and a triangle is that circles can be located inside and outside a triangle. (5 points) |Score | | | 2.
Switch out 'y' and 'x'. Now have x = 2y + 5. After you get x = 2y + 5 you want to subtract 5 from both sides, this cancels out 5 on one side. So with the
It then has a right triangle, an triangle which has a one 90-degree angle. An example of an right angle could be seen as the corner of a piece of paper. The hypotenuse will be the longest side of the triangle. “The Pythagorean Theorem states that the square of the hypotenuse in right triangle is equal to the sum of the squares of the other two sides. This is expressed as a^2 + b^2 = c^2.” (Weisstien, 2011) 13 40 Pythagorean triangles are right triangles in which all three sides are integers.
The Pythagorean Theorem states that: "The area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas of the squares upon the remaining sides." (Morris). The name is Pythagorean triple comes from the Pythagorean theorem, it states that every right triangle has side lengths fulfilling the formula a2 + b2 = c2. The Pythagorean triples describe the three integer side lengths of a right triangle, which is commonly written as (a, b, c), and a good example is (3, 4, 5). The sides of a and b are the legs of a right triangle, and c is the hypotenuse, so the formula is written as a2 + b2=c2, which satisfies what the Pythagorean Triples (Bluman, 2011).
If a point inside the figure is chosen at random, what is the probability that the point is inside the shaded region? Find the ratio of the area of the shaded region to the area of the entire semicircle. The area of a 1 semicircle is __πr 2. 2 First, find the area of the entire semicircle. 1 A t = _π(3 2) = 4.5π Total area of largest semicircle 2 Next, find the unshaded area.