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.
[2] iii) Find the equation of the circle. [2] iv) The line y + 5x = 8 cuts the circle at A and again at a second point D. Calculate the coordinates of D. [4] v) Prove that the line AB is perpendicular to the line CD. [3] 5. Find the angle ( and the length x in the triangle shown below. [7] 6.
Note: I want the direction of the conventional current. 5. Consider the circuit in Figure 3, where R1 = 5.00×102Ω, R2 = 1.00×103Ω, and VB = 10.0V . (a) What is the equivalent resistance Req of the circuit? (b) Solve the circuit.
Composition and Inverse Functions MAT 222: Intermediate Algebra Instructor October 12, 2014 Composition and Inverse Functions Functions are helpful in algebra to determine model dependencies among different variables. For example, some situations require many levels of dependencies and are useful to express the final variable in terms of the first variable. Function composition defines a variable in terms of another one while reducing the level of dependencies; and an inverse function comes into play when checking a correlation between dependent and independent variables. This paper will outline and solve a true composition and inverse function. The functions are: fx=2x+5 gx=x2-3 hx=7-x3
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.
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.
MAT 161 College Algebra Name___________________________________ Midterm Review MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Factor the polynomial by grouping. 1) x2 + 7x + 4x + 28 A) (x - 7)(x - 4) 1) B) (x - 7)(x + 4) C) (x + 7)(x + 4) D) prime Find an equation for the line with the given properties. 2) Perpendicular to the line x = 6; containing the point (5, 2) A) x = 2 B) y = 2 C) y = 5 D) x = 5 2) List the intercepts for the graph of the equation.
ALGEBRA II Regents Review 1. Simplify: = 1 2. The expressionis equivalent to which of the following? a) b) c) d) 3. Solve for x: x = 5 4.
Example (1, 2) & (3, 4) 2. Find the slope of the line connecting these two points a. Using a table with difference of y-values vs. x-values b. Using the formula m = (y2 -y1)
Putting dydx means that we need to re-arrange the limits of integration so the inside integral has limits for y, and the outside integral has the limits for x, and vice versa. In terms of the calculus 3 curriculum, this theorem makes it possible to evaluate double integrals in a much simpler way. This also makes it possible to change the limits of integration. This means looking at it horizontally versus horizontally, or vice versa. By graphing and visualizing we can