There are two types of special right triangles: 45-45-90 and 30-60-90. The legs on a 45-45-90 triangle are 1 and 1 and the hypotenuse is the square root of 2. The legs on a 30-60-90 triangle are 1 and the square root of 3 and the hypotenuse is 2. If you were to take the three trigonometric functions of either 45 degree angle, you would get the (square root of 2)/2 for both cosine (x) and sine (y) and 1 for tangent (y/x). If you were to take the three trigonometric functions of the 30 degree angle, you would get the (square root of 3)/2 for cosine, ½ for sine and the (square root of 3)/3 for tangent.
a) b) c) d) 16. Which equation represents the Objective Function for the above problem? a) b) c) d) 17. Write the equation of a square root graph that has been vertically compressed by a factor of , reflected over the x-axis, translated down 2 units and right 3 units. 18.
The diagram below shows the graph of y = –x . 2 y O x (–3, k) y = –x2 The point (–3, k) lies on the graph. Find the value of k. 1 6. C B 12 cm A 1 1 In triangle ABC, AB = 12 centimetres, sin C = 2 and sin B = 3 . Find the length of side AC.
11 11. The roots of the quadratic equationare a) real, rational, and equal b) real, rational, and unequal c) real, irrational, and unequal d) imaginary 12. Express the roots of the equationin simplest a + bi form. 13. For which value of k will the roots ofbe real?
Solve using the multiplication principle first. Then use the elimination method. 2x+y=13 4x+2y=23 8. Solve by rearranging the equations first. Then use the multiplication principle and then use the elimination method: 3x=8y+11 x+6y-8=0 9.
e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x. f) Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x. A) x² - 2x – 13 = 0 x² - 2x = 13 4x² - 8x = 52 4x² - 8x + 4 = 56 (2x – 2) ± 56 2x – 2 = 56 2x – 2 = 56 2x = 56+2 2x = 56+2 x = 56+2/2 x = 56+2/2 Project 2 C) x² + 12x – 64 = 0 -x² + 12x = 64 4x² + 48x = 256 4x² + 48 + 144 = 400 2x + 12 = ±20 2x +12 = 20 2x + 12 = -20 2x = 8 2x = -32 x = 4 x = -16 The method selected was the method shown in our text. I do not think I will ever have a chance to use this knowledge in a real-world situation. Algebra and Geometry are the hardest subjects for me, and I do not retain the teaching very well, so I am sure I will never use this formula. References Bluman, A.
We can conclude that the data are Poisson distributed. Chi-Square test of independence Problem 12.12 Use the following contingency table to determine whether variable 1 is independent of variable 2. Let α = .01 | Variable 2 | Variable1 | 24 | 13 | 47 | 58 | | 93 | 59 | 187 | 244 | Step 1 Ho: the two classifications are independent Ha: the two classifications are dependent Step 2 d.f = (r – 1) (c – 1) Step 3 α = 0.01 x 2 0.01, 3df = 11.3449 Step 4 Reject Ho if x 2 > 11.3449 | Variable 2 | Total | Variable1 | 24 (22.92) | 13 (14.10) | 47 (45.83) | 58 (59.15) | 142 | | 93 (94.08) | 59 (57.90) | 187 (188.17) | 244 (242.85) | 583
2. Verify the equation of the pendulum T = 2 q L g by means of the graph of T2 vs L and determine the value of g from the graph. 1 3 Data Collection Length (m) Period (s) 1.954 1.00 1.959 1.955 1.879 0.91 1.888 1.880 1.747 0.80 1.754 1.765 1.759 0.82 1.757 1.761 1.653 0.72 1.645 1.641 1.501 0.60 1.506 1.504 Table 1: Data Collection 4 Data Analysis See Table 2 and attached graph. Slope = 3.826 1 = 3.826 m/s2 g = (2)2 3.826 = 10.3 m/s2 2 Length L(m) Period T(s) Average T T2 1.954 1.00 1.959 1.956 3.826 1.955 1.879 0.91 1.888 1.882 3.543 1.880 1.747 0.80 1.754 1.755 3.081 1.765 1.759 0.82 1.757 1.759 3.094 1.761 1.653 0.72 1.645 1.646 2.710 1.641 1.501 0.60 1.506 1.504 2.261 1.504 Table 2: Data Analysis 5 Evaluation In this lab, we were able to estimate the value of the acceleration due to gravity g on Earth by means of a simple pendulum. Because the length of a pendulum L, and the square of the period of the pendulum T2 are directly proportional, we were able to determine g by calculating the slope of the T2 vs L graph.
n mathematics, the Pythagorean theorem—or Pythagoras' theorem—is a relation in Euclidean geometry among the three sides of a right triangle. It states that the square of thehypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The theorem can be written as an equation relating the lengths of the sides a, b and c, often called the Pythagorean equation:[1] where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides. The Pythagorean theorem is named after the Greek mathematician Pythagoras (ca. 570 BC—ca.
The letters tell you about the number of dimensions of the representation (dimensions are the characters under E on the character table). A means 1D (E = 1) and symmetric (positive value) when rotating along the principal axis. B means 1D (E = 1) and asymmetric (negative value) when rotating along the principal axis. E means 2D (E = 2) T means 3D (E = 3) 1 The subscripts and superscripts tell you about the symmetry of the reps with