algebra1

algebra1

We have to solve 2/x+3=<1/x-3 2/(x+3)=<1/(x-3) => 2/(x+3) - 1/(x-3) =< 0 => [2(x-3) - (x+3)]/(x - 3)(x + 3) =< 0 => (2x - 6 - x-3)/(x - 3)(x + 3) =< 0 => (x-9)/(x - 3)(x...

Latest answer posted February 17, 2011 12:18 am UTC

algebra1

The roots of the equation x^2+ax=2 are x1 and x2. => x1^2 + ax1 = 2 and x2^2 + ax2 = 2 Adding the two x1^2 + x2^2 + a(x1 + x2) = 4 But it is given that x1^2 + x2^2 = 4, this gives a = 0 The...

Latest answer posted February 23, 2011 1:09 am UTC

algebra1

We have to solve x+6 = 6* sqrt (x-2) x+6 = 6* sqrt (x-2) square both the sides (x + 6)^2 = 36( x - 2) => x^2 + 12x + 36 = 36x - 72 => x^2 - 24x + 108 = 0 => x^2 - 18x - 6x + 108 = 0 =>...

Latest answer posted January 20, 2011 4:47 pm UTC

algebra1

We can find the solution to this system of equations by the elimination method. Step 1: Subtract 2 from both sides of 4x + 18y +2 = 0 The equation becomes 4x + 18y = -2 Step 2: Multiply all...

Latest answer posted January 22, 2011 11:03 pm UTC

algebra1

The roots of the quadratic equation are 13 and 3. The quadratic equation can be written as (x - 13)(x - 3) = 0 (x - 13)(x - 3) = 0 => x^2 - 13x - 3x + 39 = 0 => x^2 - 16x + 39 = 0 The...

Latest answer posted February 17, 2011 12:43 am UTC

algebra1

We have E(x) = (x^3-3x-2)/(x^3+1+3x^2+3x). We have to solve E(x) = sqrt 3 - sqrt 2. E(x) = sqrt 3 - sqrt 2 => E(x) = (x^3-3x-2)/(x^3+1+3x^2+3x) = sqrt 3 - sqrt 2 => (x^3-3x-2)/(x + 1)^3 =...

Latest answer posted February 21, 2011 2:50 am UTC

algebra1

We have to solve the equation 1/2^(x^2-1) = sqrt (16^x) 1/2^(x^2-1) = sqrt (16^x) => 2^-(x^2 - 1) = 16^x^(1/2) => 2^(-x^2 + 1) = 2^4^x^(1/2) => 2^(-x^2 + 1) = 2^2x We can equate the...

Latest answer posted March 1, 2011 11:57 pm UTC

algebra1

We are given that in the quadratic equation ax^2+bx+c, a,b,c are terms of arithmetic sequence such that a = 2t-3, b = 5t+1 and c = 4t-7 As consecutive terms of an AP have a common difference: 4t -...

Latest answer posted February 17, 2011 1:30 am UTC

algebra1

We have to solve the equations: x + y = -3 ...(1) x + z = -2 ...(2) xy + yz + xz = 2 ...(3) From (1) x + y = -3 => y = -3 - x From (2) x + z = -2 => z = -2 - x Substitute these in (3) xy + yz...

Latest answer posted February 7, 2011 10:47 pm UTC

algebra1

It is given that the AP consists of the terms 1 , 6x + 2 , 13 , 19 ... As the consecutive terms of an AP have a common difference: 13 - 6x - 2 = 6x + 2 - 1 => 11 - 6x = 6x +1 => 12x = 10...

Latest answer posted February 17, 2011 1:37 am UTC

algebra1

We have to simplify sqrt 6( sqrt 3 + sqrt 12) sqrt 6( sqrt 3 + sqrt 12) => sqrt 3*2 (sqrt 3 + sqrt 3*4) => sqrt 3* sqrt 2 (sqrt 3 + 2*sqrt 3) => sqrt 3* sqrt 3* sqrt 2 (1 + 2) =>...

Latest answer posted February 23, 2011 3:04 am UTC

algebra1

As z is a complex number let it be x + yi. z' = x - yi. If z/2 + 3=z'/3 - 2 => (x + yi)/2 + 3 = (x - yi)/3 - 2 => (x + yi)/2 - (x -yi)/3 = -5 => x/2 - x/3 - i( y/2 - y/3) = -5 => x/2 -...

Latest answer posted January 21, 2011 9:31 pm UTC

algebra1

We have to solve : y^2 +3 + 12/(y^2+3)=7 y^2 +3 + 12/(y^2+3)=7 => (y^2 + 3)^2 + 12 = 7( y^2 + 3) => y^4 + 9 + 6y^2 + 12 = 7y^2 + 21 => y^4 - y^2 = 0 => y^2( y^2 - 1) = 0 y^2 = 0 or y =...

Latest answer posted February 17, 2011 1:47 am UTC

algebra1

The equation to be solved for x is 243^(x - 3) = 9^x Make the base 3 on both the sides 243 = 3^5 and 9 = 3^2 243^(x - 3) = 9^x => 3^5^(x - 3) = 3^2^x => 3^(5x - 15) = 3^2x as the base is the...

Latest answer posted May 7, 2011 10:49 am UTC

algebra1

We have to solve for x and y given that y=6/x and 2^(x+y)=32 2^(x+y)=32 => 2^(x + y) = 2^5 => x + y = 5 substitute y = 6/x x + 6/x = 5 => x^2 - 5x + 6 = 0 => x^2 - 3x - 2x + 6 = 0 =>...

Latest answer posted May 7, 2011 11:19 am UTC

algebra1

It is given that f(x)=-14x^2+14x-11 and g(x)=13x^2+13x+12 To determine f(6)/g(6) substitute x = 6 in f(6) and g(6) and divide the two results. f(6) = -14*36 + 14*6 - 11 = -431 g(6) = 13*36 +13*6 +...

Latest answer posted May 7, 2011 11:16 am UTC

algebra1

We have to solve for x and y given that 2x-y+1=0 and x^2+x-y+1=0. 2x - y + 1 = 0 => y = 2x + 1 substitute this in x^2+x-y+1=0. x^2 + x - y + 1 = 0 => x^2 + x - 2x - 1 + 1 = 0 => x^2 - x =...

Latest answer posted January 21, 2011 9:51 pm UTC

algebra1

The equation to be solved is 5*6^x = 2*3^2x + 3*2^2x 5*6^x = 2*3^2x + 3*2^2x => 5 = 2*3^2x/6^x + 3*2^2x/6^x => 5 = 2*(3/2)^x + 3*(2/3)^x let (3/2)^x = y => 5 = 2y + 3/y => 5y = 2y^2 +...

Latest answer posted May 7, 2011 11:10 am UTC

algebra1

We have the equation: 2x^3 - x^2 + ax + b = 0 with one of its solutions given as 1 + i. Substituting x with 1 + i, 2*(1 + i)^3 - (1 + i)^2 + a(1 + i) + b = 0 => 2*( 1 + 3i^2 + 3i + i^3) - 1 -...

Latest answer posted February 21, 2011 11:24 pm UTC

algebra1

x1 and x2 are given as the solutions of x^2 - x + 1 = 0 So if we substitute x with x1 we get x1^2 - x1 + 1 = 0 => x1^2 - x1 = -1 Similarly with x2, x2^2 - x2 + 1 = 0 => x2^2 - x2 = -1 The...

Latest answer posted February 21, 2011 11:28 pm UTC

algebra1

We need to simplify [(x^2+3x+2)/(x-2)]*[(x^2-4)/(x-1)] [(x^2+3x+2)/(x-2)]*[(x^2-4)/(x-1)] use x^2 - 4 = (x - 2)(x +2), also let's find the factors of x^2+3x+2 => [(x^2 + 2x + x +2)/(x-2)]*[(x -...

Latest answer posted April 14, 2011 6:42 am UTC

algebra1

For a parabola y = ax^2 + bx + c, the vertex is given by the coordinates: (-b/2a , -(b^2 - 4ac)/4a) The equation of the parabola given is : y=x^2+4x substituting the values, the vertex is (-2/2 ,...

Latest answer posted May 7, 2011 11:32 am UTC

algebra1

We have to solve the equation |6t - 9| - 21 =0 |6t - 9| - 21 =0 => |6t - 9| = 21 This gives us two equations 6t - 9 = 21 => 6t = 21 + 9 => 6t = 30 => t = 30/6 => t = 5 And 9 - 6t =...

Latest answer posted January 21, 2011 10:04 pm UTC

algebra1

We have to solve: 2^3x=2^(-x+28) 2^3x = 2^(-x + 28) we have an equal base on both the sides, so we can equate the exponential 3x = -x + 28 => 4x = 28 => x = 28/4 => x = 7 The solution is x...

Latest answer posted February 26, 2011 5:28 pm UTC

algebra1

Let 4^n - 1 represent the sum of m terms of a GP. 4^(n - 1) - 1 is the sum of m - 1 terms of the GP. Tn = 4^n - 1 - 4^(n - 1) + 1 => 4^n - 4^n/4 For two consecutive terms Tn and Tn+1 Tn+1 / Tn =...

Latest answer posted May 7, 2011 11:03 am UTC

algebra1

We have to solve 4*|8x - 8| < 32 4*|8x - 8| < 32 => |8x - 8| < 8 => |x - 1| < 1 -1 < (x - 1) < 1 -1 < (x - 1) => 0 < x (x - 1) < 1 => x < 2 The values of x...

Latest answer posted May 7, 2011 10:57 am UTC

algebra1

We have to find x for (x^1/3)^(log(x) x^2 +2)=2 log(3) 27 (x^1/3)^(log(x) x^2 +2) = 2 log(3) 27 => (x^1/3)^(log(x) x^2 + 2) = 2 log(3) 3^3 => (x^1/3)^(log(x) x^2 + 2) = 6 log(3) 3 =>...

Latest answer posted February 4, 2011 12:54 am UTC

algebra1

We have the equation y^2 + 3 = 13/y^2 - 9 and we need to find the real solutions. y^2 + 3 = 13/y^2 - 9 => y^4 + 3y^2 = 13 - 9y^2 => y^4 + 12y^2 - 13 = 0 let y^2 = x => x^2 + 12x - 13 = 0...

Latest answer posted February 26, 2011 5:35 pm UTC

algebra1

Two lines are parallel if they have the same slope. For the lines 2x-y+2=0 and x+y-4=0, the slope is : 2x - y + 2 = 0 => y = 2x + 2 slope = 2 x + y - 4 = 0 => y = -x - 4 slope = 1 The two...

Latest answer posted April 15, 2011 9:02 pm UTC

algebra1

It is given that log x^3- log 10x = log 10^5. To determine x , use the property : log a - log b = log a/b log x^3- log 10x = log 10^5 => log (x^3 / 10x) = 10^5 => x^2 / 10 = 10^5 => x^2 =...

Latest answer posted May 7, 2011 1:45 am UTC

algebra1

What we need to determine is the partial fractions of 1/(x^2+x) 1/(x^2+x) = 1/[x*(x + 1] => A/ x + B/(x + 1) => [A(x + 1) + Bx]/(x*(x + 1)) = 1/[x*(x + 1] => Ax + A + Bx = 1 equating the...

Latest answer posted April 14, 2011 6:28 am UTC

algebra1

We have the functions: v(t)=12t^2/t -1 and u(t)=t+9 If v(t) < u(t) => 12t^2/t -1 < t+9 => 12t - 1< t + 9 => 12t - t < 9 + 1 => 11t < 10 => t < 10/11 Therefore t...

Latest answer posted January 23, 2011 12:49 am UTC

algebra1

We know that f(x) = x^2 + 2x + 3 f(1) = 1^2 + 2*1 + 3 f(2) = 2^2 + 2*2 + 3 ... f(20) = 20^2 + 2*20 + 3 So the sum f(1) +f(2) ...f(20) => 1^2 + 2^2...+20^2 + 2*(1 + 2...+20) + 20 *3 =>...

Latest answer posted January 23, 2011 11:37 pm UTC

algebra1

We have to find x if (x^1/3)^(logx x^2 +2)=2log3 27 (x^1/3)^(log(x) x^2 +2) = 2 log(3) 27 => (x^1/3)^(log(x) x^2 +2) = 2*3 log(3) 3 => (x^1/3)^(log(x) x^2 +2) = 6 => (x^1/3)^(2 + 2) = 6...

Latest answer posted February 7, 2011 11:27 pm UTC

algebra1

We have to solve : 3*(4e^2x-1)^3 = 81 3*(4e^2x-1)^3 = 81 => (4e^2x-1)^3 = 27 => (4e^2x-1)^3 = 3^3 => 4e^2x - 1 = 3 => 4e^2x = 4 => e^2x = 1 => 2x = 0 as any number raised to the...

Latest answer posted March 14, 2011 5:30 pm UTC

algebra1

We have to find the factors of x^2 - 12 + 27=0. x^2 - 12 + 27=0 => x^2 + 15 = 0 Now use a^2 - b^2 = (a - b)(a + b) As 15 is positive we need to make use of i, as i^2 = -1 x^2 + 15 = x^2 -...

Latest answer posted April 14, 2011 6:34 am UTC

algebra1

It is given that 3z -9i = 8i + z + 4 3z -9i = 8i + z + 4 => 3z - z = 4 + 8i + 9i => 2z = 4 + 17i => z = 2 + 17i/2 |z| = sqrt (2^2 + (17/2)^2) => sqrt (4 + 289/4) => sqrt (305/4)...

Latest answer posted May 7, 2011 10:30 am UTC

algebra1

The segment AB is a line joining the points A( -1 , -3) and B(3, 7). The equation of the line joining them can be taken to be the function f(x). The equation of the line is given by: y + 3 = [( -3...

Latest answer posted January 23, 2011 11:46 pm UTC

algebra1

We have to solve the system x + y = 3 and x^2/y + y^2/x = 9/2 x + y = 3 => x = 3 - y Substitute this in x^2/y + y^2/x = 9/2 => (3 - y)^2 / y + y^2 / (3 - y) = 9/2 => (3 - y)(3 - y)^2 + y*...

Latest answer posted February 7, 2011 11:38 pm UTC

algebra1

We have to solve x^6 - 4x^3 + 3 = 0 Let y = x^3 x^6 - 4x^3 + 3 = 0 => y^2 - 4y + 3 = 0 => y^2 - 3y - y + 3 = 0 => y(y - 3) - 1(y - 3) = 0 => (y - 1)(y-3) = 0 We get y = 1 and y = 3 Now...

Latest answer posted January 31, 2011 9:18 pm UTC

algebra1

We have to solve the equation: 2*ln x - 4*ln 11 = 0 We know that n*log a = log a^n 2*ln x - 4*ln 11 = 0 => ln x - 2*ln 11 = 0 => ln x = 2*ln 11 => ln x = ln 11^2 taking the antilog of both...

Latest answer posted February 22, 2011 12:23 am UTC

algebra1

We have to solve for x given x^3 + 3 = 0 x^3 + 3 = 0 => x^3 = -3 => x = (-3)^(1/3) => -1.4422( approximately) x = -3^(1/3).

Latest answer posted January 31, 2011 9:21 pm UTC

algebra1

We have to solve the following system of equations by substitution. x^2 - y^2 = 9 x - y = 1 x - y = 1 => x = 1 + y Substitute this for x in x^2 - y^2 = 9 => (1 + y)^2 - y^2 = 9 => 1 + y^2...

Latest answer posted February 4, 2011 1:34 am UTC

algebra1

A fraction (4x^3-32)/[x^3+(x+2)^3] is not defined whenever the denominator is equal to 0. x^3+(x+2)^3 = 0 => x^3 + x^3 + 12x + 6x^2 + 8 = 0 => 2x^3 + 6x^2 + 12x + 8 = 0 => x^3 + 3x^2 + 6x...

Latest answer posted February 11, 2011 1:18 am UTC

algebra1

We have to solve for x given: 15^(2x-3)=3^x*5^(3x-6) 15^(2x-3)=3^x*5^(3x-6) => (5*3)^(2x - 3) = 3^x * 5^(3x - 6) => 5^(2x - 3) * 3^(2x - 3) = 3^x * 5^(2x - 3)* 5^x / 5^3 => 3^(2x - 3) =...

Latest answer posted February 18, 2011 2:34 am UTC

algebra1

The required line is perpendicular to -2x + 4y - 1 = 0 -2x + 4y - 1 = 0 => 4y = 2x + 1 => y = x/2 + 1/4 This is of the form y = mx + c where the slope is m and the y-intercept is c. Therefore...

Latest answer posted January 24, 2011 12:07 am UTC

algebra1

a1 , a2 , a3 , a4 and a5 are terms of a GP. So each term is equal to the previus term in the series multiplied by a common ratio. As log(3) a1 + log(3) a2 +...+ log(3) a5 =2, =>log(3)...

Latest answer posted February 4, 2011 1:57 am UTC

algebra1

We have to verify if (a+b)^5 - a^5 - b^5 = 5ab(a+b)(a^2+ab+b^2) We can expand (a + b)^5 = a^5 + 5a^4b + 10 a^3b^2 + 10 a^2b^3 + 5ab^4 + b^5 The left hand side (a+b)^5 - a^5 - b^5 => a^5 + 5a^4b...

Latest answer posted February 8, 2011 12:28 am UTC

algebra1

We have to solve the simultaneous equation y^2 - 4xy + 4y = 1 ...(1) 3x^2 - 2xy = 1 ...(2) We see that it is not possible to isolate x in terms of y or y in terms of x using any of the equations....

Latest answer posted February 8, 2011 12:47 am UTC