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Download Please show work. 1. (y-1)(y-7)=0 Hence, either of the factors can be
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Please show work. 1. (y-1)(y-7)=0 Hence, either of the factors can be zero. If y-1=0, we will have y =1. If y-7=0, we will have y =7. Hence, y =1 and y=7. 2. 25n2=4 Or, n2= 4/25 Or, n =±(2/5) 3. x2-2x-24=0 Or, x2-6x+4x-24=0 Or, x(x-6)+4(x-6)=0 Or, (x-6)(x+4)=0 Or, x = -4,6 4. We have 235 = -15t2+120t+10 Or, 15t2-120t+225=0 Or, t2-8t+15=0 Or, t2-5t-3t+15=0 Or, t(t-5)-3(t-5)=0 Or, (t-3)(t-5)=0 Or, t = 3,5 5. If w is width, length = w+5, hence, w(w+5)=66. Or, w2+5w-66=0 Or, w2-6w+11w-66=0 Or, w(w-6)+11(w-6)=0 Or, (w-6)(w+11)=0 Or,w = 6, -11 Neglecting –ve value, we get width = 6 ft and length = 6+5= 11ft. 6. We can assign any value of x from -∞ to +∞. Hence, domain: (-∞,∞) and A is correct alternative. 7. We can assign any value of x except ±2. Hence, we can write Domain:{x|-∞≤x<-2U-2<x<2U2<x≤∞} or, [-∞,-2),(-2,2),(2,∞] 8. X F(x) 0 1 2 3 4 -2 ∞ 2 1 2/3 9. (a) (f+g)(x) = x+5+x+9=2x+14 (b) (f-g)(x)=x+5-x-9= -4 (c) (fg)(x) = (x+5)(x+9)=x2+14x+36 (d)(f/g)(x) = (x+5)/(x+9) 10. Vertex:(3,-10) 11. We have f(-3)=-6 and f(-1)=-2. 12. Minimum value of y is -9. 13. (a) A is correct alternative. (b) Vertex:(-4,0) (c) Graph of y=x2 is shifted to right by 4 units with respect to the graph of y=(x+4)2. 14. y=-(x+3)2+5 15. Or, 3x2+11x-34=0 Or, 3x2+6x-17x-34=0 Or, 3x(x+2)-17(x+2)=0 Or, (3x-17)(x+2)=0 Or, x = 17/3, -2 16. Square root od 1st term = x, double it:2x and dividing 21x by this, we get 21x/2x = 21/2. Square it: 441/4, hence, 441/4 should be added. We will have trinomial: x2+21x+441/4 = (x+21/2)(x+21/2) 17. 2x2+20x-22=0 Or, x2+10x-11=0 Or, x2+11x-x-11=0 Or, x(x+11)-1(x+11)=0 Or, (x-1)(x+11)=0 Or, x =1,-11 18. s 1 2 15s 15s 1 gt t 2 t 15sg 15 g g g 19.A y 3 x 2 3 x 3 3 x 2 3 x 3 0 for y int ercept. 3 9 36 3 3 5 1 5 6 6 2 1 5 1 5 Hence, y int ercept : , 2 2 x 20. We have y = 0 at x =6 and x=8; hence, B is correct alternative. 21. Domain:{x|7≤x≤∞} 22. Or, x2+10x=-6 Square root od 1st term = x, double it:2x and dividing 10x by this, we get 10x/2x =5. Square it: 25, hence, 25 should be added on both sides. 23.Yes, it is polynomial fuction. Degree: 1 Type: linear 24. Yes, it is polynomial with degree 1. 25. Yes, it is polynomial with degree 3. Out of the above problems which one was the hardest? AND Why do you think they were more difficult than others? What did you do to overcome this difficulty? How did you arrive at a solution? Solving 2x2+20x=22 was the hardest. It was so as finding factors of 44 such that their difference 20 was difficult. I found all the factors and manipulated to get factors such that their difference 20. I reduced it to x2+10x-11=0 and factors of 11 are 11 and 1 whose difference is 10 and solved the equation. -Other than those listed in the text, how might the Pythagorean theorem be used in everyday life? Provide example of each. Pythagorean Theorem can be used to determine corner to corner distance of a rectangular table. It can be used to determine any side of right angled triangle if any two other sides are known. -Is the compound interest formula – such as would be used to calculate a car loan – an example of a function? If yes, of what type of function is it an example? Why? A = P(1+i)^n. It is polynomial function depending on value of n. -What is the relationship between exponents and logarithms? How would you distinguish between the two using both a graph and a sequence? We have 102=100 in exponent form. Or, log10100=2 is in logarithm form. Log of any –ve number is not defined. Logarithmic graphs use logarithmic scales, in which the values differ exponentially. For example, instead of including marks at 0, 1, 2, and 3, a logarithmic scale may include marks at 0.1, 1, 10, and 100, each an equal distance from the previous and next. -Explain the four steps for solving quadratic equations. Can any of these steps be eliminated? Can the order of these steps be changed? Would you add any steps to make solving quadratic equations easier or easier to understand? Let the equation to be solved is ax2+bx+c=0. The steps are A. Move the constant term to the right side of the equation. B. Multiply each term in the equation by four times the coefficient of the x^2 term. C. Square the coefficient of the original x term and add it to both sides of the equation. D. Take the square root of both sides and set the left side of the equation equal to the square root of the number on the right side of the equation and solve for x. In the four steps, we are keeping unknown terms on LHS. After this, we are making whole square on LHS. Order can be changed but it will complicate the solution. We should try to make whole square on LHS, if possible. If it is possible to factorize left side, it will be easy to solve the equation.