How do I solve this math question?
First let's try to develop a general theory for these kinds of questions. So lets consider a general equation of Conic section which is [math]ax^2+2hxy+by^2+2fx+2gy+c = 0[/math]Rearrange this equation as [math]ax^2+2x(hy+f)+(by^2+2gy+c) = 0 [/math]Solving for the [math] x[/math] we have [math]x= \dfrac{-2(hy+f)\pm2\sqrt{(hy+f)^2-a(by^2+2gy+c)}}{2a} [/math]Now if we want that given equation should have linear factors then this also means that above expression should have linear relationship between [math]x[/math] and [math]y[/math]. Now for this to happen the term under square-root should be a perfect square.[math]\implies (hy+f)^2-a(by^2+2gy+c)[/math] should be perfect square Simplifying this equation and we will get [math](h^2-ab)y^2+2y(gh-af)+(g^2-ac) = 0 [/math]Now this equation will be a perfect square if it has only one root i.e it's discriminant is zero.[math]\implies 4(gh-af)^2 - 4(g^2-ac)(h^2-ab) = 0[/math] Simplify this and we will get [math]\boxed{af^2+ch^2+bg^2-2fgh-abc = 0}[/math] Hence this condition should be true if you want to factorize any such equation into two linear factor. More neat and aesthetic form to express this condition is [math] \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 [/math]. This is also called discriminant of Conic section . (It decides whether the given equation represent an ellipse or parabola or circle or point or hyperbola or pair of straight line. Here it represents the pair of straight line).Now come to this particular equation. (I am changing [math]a[/math] to [math]P[/math] and [math]b[/math] to [math]Q[/math] to avoid confusion).So given equation is [math]3x^2 + 2Qxy + 2y^2 + 2Px - 4y + 1[/math]. If you compare this to standard form you will find that ,[math]a= 3 , h = Q , g = P , b = 2 , f = -2[/math] and [math]c = 1[/math] So discriminant of this equation is [math] \begin{vmatrix} 3 & P & Q \\ Q & 2 & -2 \\ P & -2 & 1 \end{vmatrix} = 0[/math]Solving and rearranging this we have ,[math]Q^2+4PQ + 2P^2 + 6 = 0 [/math]Clearly [math]q[/math] is a root of quadratic equation given by [math]x^2+4Px+2P^2+6[/math]
Solve 5x-2=3x-8 and represent the solution. On a number line In the cartesian plane?
5x - 2 = 3x - 8 5x - 3x = -8 + 2 2x = -6 x = -3 On a number line in the cartesian plane x is a line below the x axis at a distance of 3 units.
PLEASE HELP OMG. HURRY PLEASE I NEED HELP?
assuming monthly compounding, interest rate = 9.4/1200 = 47/6000 discount rate = 1/(1 +47/6000) = 6000/6047 = D(say) period = 12*6 = 72 let monthly payments be R, then 9640 = RD + RD^2 + RD^3 + ..... RD^72, a GP with a = RD, r = D, sum = RD(1-D^72)/(1-D) plugging in, R = 9640*(47/6047)/((6000/6047)*(1- (6000/6047)^72)) = $175.69 so total amount paid = 72*175.69 + 1175 = $13,824.68 [ A ] <-----
Need Help. What is the slope of the given line by the equation below?
6/7
Math help please hurry!!!!!!!!!??
3) Put both equations into y = mx + b form 4x – 8y = 0 -8y = -4x + 0 y = (1/2)x –3x + 8y = –7 8y = 3x - 7 y = (3/8)x - 7/8 The lines are neither parallel nor perpendicular so there is one solution set (second choice) 4) 2x – y = 5 or y = 2x - 5 4x + y = 7 or y = -4x + 7 combine: 2x - 5 = -4x + 7 6x = 12 x = 12/6 = 2 using first equation: y = 2(2) - 5 = -1 solution (2, -1) (third choice) 5/6) c + a = 33 or a = 33 - c 3c + 5a = 115 substitute first equation into second equation: 3c + 5(33 - c) = 115 3c + 165 - 5c = 115 -2c = -50 c = -50/-2 = 25 using first equation: a = 33 - 25 = 8 check using second equation: 3(25) + 5(8) = 115 75 + 40 = 115 checks There were 8 adult tickets and 25 children's tickets sold. (fourth choice) 7) Second set has no common solution. 8) First set of equations is correct. 9) d + e = 6 d – e = 4 or d = e + 4 substitute second equation into first equation: e + 4 + e = 6 2e = 2 e = 2/2 = 1 using second equation: d = 1 + 4 = 5 solution (5, 1) (second choice) 10) second choice 11) 5x + 2y = 14 y = x – 7 substituting: 5x + 2(x - 7) = 14 5x + 2x - 14 = 14 7x = 28 x = 28/7 = 4 using second equation: y = 4 – 7 = -3 solution (4, -3) (third choice) - .--
Algebra II Help? 10 points. Multiple Choice. Hurry, Please?
8. What is the range of the data set? 8 34 23 50 25 16 48 32 56 42 48 3. What is the standard deviation of the data set? 44 80 63 50 35 63 71 6.4 13.4 14.6 15.8
Math help please hurry!!! multiple choice !?
1) 5x + y = 2 or y = -5x + 2 4x + y = 4 or y = -4x + 4 combine: -5x + 2 = -4x + 4 -x = 2 x = -2 using first equation: y = -5(-2) + 2 = 12 x = -2, y = 12 (B) 2) I don't have a working scanner so I'll solve mathematically. x – y = 3 or x = y + 3 x + y = –3 or x = -y - 3 combine: y + 3 = -y - 3 2y = -6 y = -6/2 = -3 using first equation: x = -3 + 3 = 0 solution (0, -3) (A) 3) 2x + 6y = –12 or y = (-1/3)x - 2 10x + 32y = –62 or y = (-5/16)x - 31/21 There is one solution (third choice) 4) 2x + 6y = 4 or y = (-1/3)x + 2/3 5x + 15y = 10 or y = (-1/3)x + 2/3 there are an infinite number of solutions as the lines are congruent (fourth choice) 5) 3d – e = 7 or e = 3d - 7 d + e = 5 or e = -d + 5 combine: 3d - 7 = -d + 5 4d = 12 d = 12/4 = 3 using first equation: e = 3(3) - 7 = 2 solution (3, 2) (first choice) - .--
Pre Algebra Help! HURRY 10 POINTS GIVEN?
1. Which equation can be used to find the perimeter of a regular octagon with sides of length 12? (Points: 1) P = 8 + 12 P = 8(12) P = 12 ÷ 8 P = 2(8) + 2(12) 2. What is the perimeter of a rectangle with a width of 23.6 cm and a length of 52.9 cm? (Points: 1) 76.5 cm 100.1 cm 153 cm 180 cm 3. Three sides of a pentagon have a length of 26 cm each. Each of the remaining two sides has a length of 14.5 cm. What is the perimeter of the pentagon? (Points: 1) 75 cm 81 cm 107 cm 377 cm