# See tutors' answers!

Algebra ->  Tutoring on algebra.com -> See tutors' answers!      Log On

 Tutoring Home For Students Tools for Tutors Our Tutors Register Recently Solved
 By Tutor
| By Problem Number |

Tutor:

# Recent problems solved by 'tam144'

tam144 answered: 5 problems
 Equations/667821: 4x-3y<81 solutions Answer 415199 by tam144(5)   on 2012-10-17 09:58:50 (Show Source): You can put this solution on YOUR website!The best way is to solve it by graphing. 4x - 3y - 8 < 0 (1) First graph the linear function: f(x,y) = 4x - 3y - 8 = 0. When x = 0 ---> y = -8/3 When y = 0 ---> x = 2. The graph is the line passing at 2 points A(0, -8/3) and B(2, 0). Now use the check point method to find the area that is the solution set. Check with the origin O. Substitute x = 0 and y = 0 into the inequality (1). We get -8 < 0. It is true, then the area containing the origin is the solution set. Re-check with the point (3, 0). Substitute y = 0 and x = 3 into the inequality (1). We get: 12 - 8 < 0. It is not true, then the other area is the answer.
 Trigonometry-basics/667155: Would be ever so much grateful for help to prove the trigonometric relation in a thoroughly detailed explanation of proof. 1 solutions Answer 414843 by tam144(5)   on 2012-10-16 08:35:47 (Show Source): You can put this solution on YOUR website!Start from the right side of the equation. Refer to List of Trig Identities in book titled:"Solving trig equations and inequalities" (Amazon e-book 2010) 1 - cos x = 2 sin^2 x/2 (Identity 11B) sin x = 2sin x/2.cos x/2 (Identity 10) Quotient: (1 - cos x)/sin x = (2sin^2 x/2)/(2sin x/2.cos/x/2) After simplification, the quotient becomes: !1 - cos x)/sin x = (sin x/2) / (cos x/2) = tg x/2
 test/639002: Please help me solve this equation: this is factor completely1 solutions Answer 402537 by tam144(5)   on 2012-08-20 10:06:55 (Show Source): You can put this solution on YOUR website!Solve x^2 + 8x - 9 = 0. There are 2 Tips you should know when you meet this special type of quadratic equation. The standard form of a quadratic equation is: ax^2 + bx + c = 0. Tip 1. When a + b + c = 0, one real root is x1 = 1 and the other x2 = c/a. Example 1. Solve x^2 + 8x - 9 = 0. You see that a + b + c = 1 + 8 - 9 = 0. This is Tip 1. One real root is x1 = 1 and the other x2 = c/a = -9/1 = -9. Example 2. Solve 7x^2 - 15x + 8 = 0. You find 7 - 15 + 8 = 0. This is Tip 1. The 2 real roots are: x1 = 1 and x2 = c/a = 8/7. TIP 2. When a - b + c = 1, one real root x1 = -1 and the other x2 = -c/a. Example 3. Solve 3x^2 - 5x - 8 = 0. You find 3 -(-5) - 8 = 0. This is Tip 2. Then one real root x1 = -1 and the other x2 = -c/a = 8/1 = 8. Example 4. Solve 5x^2 + 17x + 12 = 0 . You find 5 -(17) + 12 = 0. Then x1 = -1 and x2 = -c/a = -12/5. I advise you to remember these 2 Tips. It will save you a lot of time in solving quadratic equations. Or, you can solve the given equation x^2 + 8x - 9 = 0 by the popular factoring method. Find 2 numbers whose product is -9 and whose sum is 8. Proceed: (-1, 9),(1, -9)OK. Factor the equation: x^2 + 8x - 9 = (x - 1)(x + 9) = 0. Next, solve the 2 binomials for x. x - 1 = 0 --> x = 1 and (x + 9) = 0 --> x = -9.
 Equations/360016: Hi, I'm studying for a test and having trouble solving the below problem: 2x^2 + 3x < 9 I'm assuming/hoping this is a completing the square problem? x^2 + 3/2x < 9/2 x^2 + 3/2x + 9/4 < 27/4 This is where my head starts to spin lol. I'm supposed to find the square of the quadratic on the left...there has got to be an easier way to do this! If not, then I'm really not sure how I would go about factoring that equation, so any guidance you can provide is greatly appreciated!1 solutions Answer 401319 by tam144(5)   on 2012-08-13 00:06:54 (Show Source): You can put this solution on YOUR website!Always solve a quadratic inequality in 4 steps. Step 1. Bring it to standard form: f(x) = 2x^2 + 3x - 9 < 0. Step 2. Solve f(x) = 0. You can use any method you prefer. I use the new Diagonal Sum method. Roots have opposite signs. There are 3 probable root-pairs: (-1/2, 9/1),(-3/1, 3/2),(-3/2, 3/1). The diagonal sum of the second set is -3 = -b. The 2 real roots are -3 and 3/2. Or, you can solve it by the factoring ac method (You Tube). Find 2 number that their product is ac = -18, and their sum is b = 3. Proceed: [(-1, 18)(1, -18)(-2, 9)(2, -9)(-3, 6), OK]. Replace in the equation f(x) = 0 the quantity 3x by two quantities -3x and 6x. 2x^2 + 3x - 9 = 2x^2 - 3x + 6x - 9 = 0. = 2x(x + 3)- 3(x +3) = (x + 3)(2x - 3). Solve the 2 binomials: x + 3 = 0 ---> x = -3 2x - 3 = 0 ---> x = 3/2 Step 3. Solve the inequality f(x) < 0. Use the number line and test point method. Plot the 2 real roots -3 and 3/2 on the number line. Use the origin O as test point. Substitute x = 0 into the inequality. You get -9 < 0. It is true, then the origin O is on the true segment (-3, 3/2). Step 4. Express the answer (solution set) of the inequality in the form of an open interval (-3, 3/2). The 2 end points -3 and 3/2 are not included in the solution set. If in the inequality, there is an additional (=) sign (lesser or equal to), then the solution set is a closed interval [-3, 3/2]. The 2 end points -3 and 3/2 are included in the solution set.
 Trigonometry-basics/614417: 1 solutions Answer 386544 by tam144(5)   on 2012-05-24 09:24:59 (Show Source): You can put this solution on YOUR website!Problem: cos x = 2/3 ; tan x < 0 ; Find exact values of sin x. Proposed solution. Suppose a calculator gives cos a1(deg.) = 2/3. On the trig unit circle, there are 2 arcs that have the same cos value (2/3). They are a1(deg.) and a2 = (360 - a1). The tan a2 is negative (< 0). The exact value of sin a2 is given by the trig identity: sin^2 a2 = 1 - cos^2 a2 = 1 - 4/9 = 5/9. There are 2 answers: sin a2 = 2.24/3 = 0.75 and sin a2 = -0.75, but only the negative value (-0.75) is the right answer.