algebra1

algebra1

We have to solve for x: log2 x + log4 x + log8 x =11/6 log (a) b = 1/ log (b) a log2 x + log4 x + log8 x =11/6 => 1 / log(x) 2 + 1 / log(x) 4 + 1/log(x) 2 = 11/6 => 1 / log(x) 2 + 1 /...

Latest answer posted January 21, 2011, 9:07 pm (UTC)

algebra1

k! = 1*2*3*...*k (k - 1)/k! = k/k! - 1/k! => 1/(k - 1)! - 1/k! The sum of (k - 1)/k! for k = 1 to n is: 1/0! - 1/1! + 1/1! - 1/2! + 1/2! - 1/3! + ... + 1/(n - 1)! - 1/n! => 1/0! - 1/n!...

Latest answer posted June 2, 2011, 12:26 am (UTC)

algebra1

The roots of the quadratic equation are x1 and x2. |x1 - x2| = 1 and x^2 = 2x - m As x^2 = 2x - m x1^2 = 2* x1 - m x2^2 = 2*x2 - m Subtracting the two we get x1^2 - x2^2 = 2( x1 - x2) => (x1 -...

Latest answer posted January 26, 2011, 5:11 pm (UTC)

algebra1

We have x^2 + y^2 = 100 and x - y = 2. We have to solve these equations for x and y. x - y = 2 => (x - y)^2 =2^2 => x^2 + y^2 - 2xy = 4 substitute x^2 + y^2 = 100 => 100 - 2xy = 4 =>...

Latest answer posted February 23, 2011, 12:50 am (UTC)

algebra1

The nth term of an arithmetic sequence can be written as a + (n - 1)*d, where a is the first term and d is the common difference. We have a2-a6+a4=-7 => a + d - a - 5d + a + 3d = -7 => a - d...

Latest answer posted February 21, 2011, 1:25 am (UTC)

algebra1

The multiplicative inverse is another name for the reciprocal. Both numbers are such that, if multiplied by the given number, they result in 1. The number 1 is special with regard to the operation...

Latest answer posted September 30, 2018, 6:37 pm (UTC)

algebra1

The roots of x^3 - 9x^2 + 23x - 15 = 0 are in AP. x^3 - 9x^2 + 23x - 15 = 0 => x^3 - 3x^2 - 6x^2 + 18x + 5x - 15 = 0 => x^2(x -3 ) - 6x ( x - 3) + 5(x - 3) =0 => (x^2 - 6x + 5)(x - 3) = 0...

Latest answer posted March 3, 2011, 1:31 am (UTC)

algebra1

We need to solve (log(2) x)^2 + log(2) (4x) = 4 Use the property that log a*b = log a + log b (log(2) x)^2 + log(2) (4x) = 4 => (log(2) x)^2 + log(2) 4 + log(2) x = 4 => (log(2) x)^2 + 2 +...

Latest answer posted May 3, 2011, 12:50 pm (UTC)

algebra1

The roots of a quadratic equation are given as 6 and 7. This means that x - 6 = 0 and x - 7 = 0 We can write ( x - 6)( x - 7) = 0 => x^2 - 6x - 7x + 42 = 0 => x^2 - 13x + 42 = 0 Therefore the...

Latest answer posted January 22, 2011, 3:17 am (UTC)

algebra1

x^3 - y^3 = 7..........(1) x^2 + xy + y^2 = 7 ..............(2) We need to solve the system. First we will rewrite equation (1) as a difference between cubes. ==> x^3 - y^2 = (x-y)(x^2 +...

Latest answer posted February 4, 2011, 3:37 am (UTC)

algebra1

We have to find the zeros of the function f(x) = x^2 - 8x - 9 For this we equate f(x) to 0 and solve the quadratic equation that is obtained. f(x) = 0 => x^2 - 8x - 9 = 0 => x^2 - 9x + x - 9...

Latest answer posted February 22, 2011, 12:34 am (UTC)

algebra1

We have to solve the following set of simultaneous equations: 2x - 3y = 5 ...(1) x - 2y = 4 ...(2) From (2) x - 2y = 4 => x= 4 + 2y Substitute in (1) 2( 4 + 2y) - 3y = 5 => 8 + 4y - 3y = 5...

Latest answer posted February 12, 2011, 10:15 pm (UTC)

algebra1

We have to solve the equation sqrt (x + sqrt (1 - x)) + sqrt x = 1 sqrt (x + sqrt (1 - x)) + sqrt x = 1 sqrt (x + sqrt (1 - x)) = 1 - sqrt x square both the sides => x + sqrt(1 - x) = 1 + x - 2...

Latest answer posted February 23, 2011, 12:59 am (UTC)

algebra1

We have to determine the imaginary part of the complex number z for z^2 = 3 + 4i Let z = x + yi z^2 = x^2 + y^2*i^2 + 2xyi = 3+ 4i => x^2 - y^2 + 2xyi = 3 + 4i equate the real and imaginary...

Latest answer posted January 29, 2011, 2:45 am (UTC)

algebra1

The multiplicative inverse of an element of a set is another element of this set such that the product of the two elements is a multiplicative identity. In the given set of complex numbers, the...

Latest answer posted January 13, 2019, 12:06 am (UTC)

algebra1

The polar form of a complex number z = x + yi is r*(cos A + i* sin A) where tan A= y/x and r = sqrt (x^2 + y^2) Here we have z = 6 - 8i r = sqrt( 6^2 + 8^2) = sqrt (36 + 64) = sqrt 100 = 10 tan A =...

Latest answer posted March 3, 2011, 5:31 pm (UTC)

algebra1

Let the first term of the arithmetic sequence be a and the common difference be d. (a + a + 12d)*(13/2) = 130 => 2a + 12d = 20 => a + 6d = 10 => a = 10 - 6d a4, a10 and a7 are consecutive...

Latest answer posted February 8, 2011, 12:21 am (UTC)

algebra1

Consecutive terms of a GP have a common ratio. if 2, x, y, 16 form a GP. => 16/y = x/2 => x = 32/y y/x = x/2 Substitute x = 32/y => y/(32/y) = (32/y)/2 => 2y = (32/y)^2 => 2y^3 =...

Latest answer posted May 11, 2011, 5:23 am (UTC)

algebra1

In this question, I am not clear whether the logs have the base of 3 or 3 is the part of the arguments. I will try to show both cases. Let's consider the given system of equations: 4^(x/y)*4^(y/x)...

Latest answer posted November 21, 2018, 4:42 pm (UTC)

algebra1

We have the system of equations x^2 + y^2 = 16 and xy = 13 to solve for x. xy = 3 => x = 3/y...(1) Substitute this in x^2 + y^2 = 16 => (3/y)^2 + y^2 = 16 => 9 + y^4 = 16y^2 => y^4 -...

Latest answer posted April 30, 2011, 12:42 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 domain of a function f(x) is all the values x for which f(x) gives real values. f(x)=(x-2)/(x^2-4) => (x - 2)/(x - 2)(x + 2) => 1/(x + 2) This is not defined when x = -2 The domain is R -...

Latest answer posted May 7, 2011, 11:24 am (UTC)

algebra1

We are given that a1 + a2 + a3 + ... + an = (5n^2+6n). Sn = a1 + a2 + a3 + ... + an = (5n^2+6n). Sn+1 = a1 + a2 + a3 + ... + an + an+1 = (5(n+1)^2+6(n+1)). Sn+1 - Sn => a1 + a2 + a3 + ... + an +...

Latest answer posted February 22, 2011, 1:02 am (UTC)

algebra1

We have to find the real solutions of log(x+2) x + log(x) (x+2) = 5/2. We use the relation log (a) b = 1/log (b) a and log a + log b = log a*b log(x+2) x + log(x) (x+2) = 5/2 => 1/ log (x) (x+2)...

Latest answer posted March 3, 2011, 5:43 pm (UTC)

algebra1

We have to solve 2z + 6i = z/2i + 5i - 7 2z + 6i = z/2i + 5i - 7 => 2z - z/ 2i = -i - 7 => (2z*2i - z) / 2i = (-i - 7) => z( 4i - 1)/2i = (-i - 7) => z = (-i - 7) * 2i / ( 4i - 1) =>...

Latest answer posted February 18, 2011, 7:35 pm (UTC)

algebra1

The sum of 3+2i+1-5i is found by adding the real terms together and the complex terms, or the terms that contain i, together. 3+2i+1-5i => 3 + 1 + 2i - 5i => 4 - 3i The required result of...

Latest answer posted April 14, 2011, 6:30 am (UTC)

algebra1

We have to solve: 5(8e^2x - 3)^3 = 625 for x. 5(8e^2x - 3)^3 = 625 divide both sides by 5 => (8e^2x - 3)^3 = 125 take the cube root of both the sides => 8e^2x - 3 = 5 add 3 to both the sides...

Latest answer posted March 11, 2011, 2:19 am (UTC)

algebra1

We have to solve: e^2x - 6e^x + 5 = 0 e^2x - 6e^x + 5 = 0 let e^x = t => t^2 - 6t + 5 = 0 => t^2 - 5t - t + 5 = 0 => t(t - 5) - 1( t - 5) = 0 => (t - 1)(t - 5) = 0 => e^x = 1 and e^x...

Latest answer posted March 11, 2011, 2:25 am (UTC)

algebra1

The complex number (2+2i)^11/(2-2i)^9 has to be simplified first. (2+2i)^11/(2-2i)^9 => (4 + 4i^2 + 8i)^5*(2 + 2i) / (4 +4i^2 - 8i)^4 * (2 - 2i) => (8i)^5*(2 + 2i) / (- 8i)^4 * (2 - 2i) =>...

Latest answer posted April 18, 2011, 12:34 am (UTC)

algebra1

We see that for each value of x, f(x)=(2x-5)/(7x+4) has only one value and each value of f(x) can be obtained by only one value of x. The function has an inverse. Let y = f(x)=(2x-5)/(7x+4) express...

Latest answer posted May 1, 2011, 11:23 am (UTC)

algebra1

We have to solve the simultaneous equations x^2 + y^2=10 ...(1) x^4 + y^4=82 ...(2) let a = x^2 and b = y^2 This gives a + b = 10 and a^2 + b^2 = 82 a + b = 10 => a^2 + b^2 + 2ab = 100...

Latest answer posted February 23, 2011, 1:36 am (UTC)

algebra1

We are given that f(x)= 6^x and g(x) = log(6) x We have to find fog(36) fog(36) = f(g(36)) => f(log(6) 36) => 6^(log (6) 36) we know that a^(log(a) x) = x => 36 The required solution for...

Latest answer posted March 8, 2011, 12:52 am (UTC)

algebra1

A polynomial has complex roots in pairs of conjugates. As the polynomial has roots 2 and 2i, it also has -2i as a root. The polynomial is: (x - 2)(x - 2i)(x + 2i) => (x - 2)(x^2 - 4i^2) => (x...

Latest answer posted May 11, 2011, 5:36 am (UTC)

algebra1

The terms x + 4, 3x and 13x - 2 are consecutive terms of an arithmetic progression. So we know that they have a common difference 13x - 2 - 3x = 3x - x - 4 => 10x - 2 = 2x - 4 => 8x = -2...

Latest answer posted January 27, 2011, 3:40 am (UTC)

algebra1

The polynomial ax^3 + bx^2 + cx + d is the product of the terms (x-1), (x+1) and (x+2) to which the remainder 3 is added. (x - 1)(x +1)(x + 2) +3 => (x^2 - 1)(x + 2) +3 => x^3 - x + 2x^2 - 2...

Latest answer posted January 31, 2011, 4:18 pm (UTC)

algebra1

We have to solve for x given that x^log2 x + 8*x^-log2 x = 6 x^log(2) x + 8*x^(- log(2) x) = 6 => x^log(2) x + 8/x^(log(2) x) = 6 Let x^log(2) x = y => y + 8/y = 6 => y^2 - 6x + 8 = 0...

Latest answer posted February 4, 2011, 12:40 am (UTC)

algebra1

The first step in finding an inverse function is to write the function in “y equals” notation, as it will allow us simply to switch the variables and solve for us once again as follows: Let...

Latest answer posted November 28, 2018, 12:51 pm (UTC)

algebra1

The equation you have given is z^4 - 3z^2 + 2 = 0. Here you can replace z^2 = x to get a quadratic equation in x which can be solved. z^4 - 3z^2 + 2 = 0 => x^2 - 3x + 2 = 0 => x^2 - 2x - x +...

Latest answer posted January 20, 2011, 11:42 pm (UTC)

algebra1

We are given that x^3 - 1728 = 0 and we have to find x. x^3 - 1728 = 0 => x^3 = 1728 => x^3 = 12^3 As the exponent is the same, we can equate the bases This gives x = 12.

Latest answer posted February 25, 2011, 10:51 pm (UTC)

algebra1

To simplify (8-6i)(-4-4i) open the brackets and multiply the terms (8-6i)(-4-4i) => 8*-4 - 8*4i - 6i*(-4) - 6i*(-4i) => -32 - 32i + 24i + 24i^2 => -32 - 8i - 24 => -56 - 8i The required...

Latest answer posted May 7, 2011, 11:13 am (UTC)

algebra1

To solve 2x - y = 5 ...(1) 3x - 2y = 9 ...(2) using substitution take (1) 2x - y = 5 => y = 2x - 5 substitute in (2) 3x - 2(2x - 5) = 9 => 3x - 4x + 10 = 9 => -x = -1 => x = 1 y = 2x -...

Latest answer posted May 11, 2011, 5:44 am (UTC)

algebra1

It is given that f(x)=14x+13. f(f(4)) given that f(4) = 14*4 + 13 => f( 14*4 + 13) => f(69) => 14*69 + 13 => 979 The value of f(f(4)) = 979

Latest answer posted May 7, 2011, 11:05 am (UTC)

algebra1

The extreme of the parabola is determined by finding the first derivative and equating it to 0. We then solve for x. y = mx^2+2mx+x+m-1 y' = 2mx + 2m + 1 = 0 => 2mx = -2m - 1 => x = (-2m - 1)...

Latest answer posted January 23, 2011, 1:15 am (UTC)

algebra1

Using the remainder theorem, as x = 1 is a root of the expression 5x^3-4x^2+7x-8=0, the expression is divisible by ( x - 1). (ax^2 + bx + c)(x - 1) = 5x^3-4x^2+7x-8 => ax^3 + bx^2 + cx - ax^2 -...

Latest answer posted January 24, 2011, 2:52 am (UTC)

algebra1

For an AP the nth terms can be written as a + (n-1)*d, where a is the first term and d is the common difference between consecutive terms. The sum of the first n terms is (t1 + tn)*(n/2) In the...

Latest answer posted February 14, 2011, 9:06 pm (UTC)

algebra1

We have to find the complex number z given that (3z - 2z')/6 = -5 Let z = a + ib, z' = a - ib (3z - 2z')/6 = -5 =>(3(a + ib) - 2(a - ib)) = -30 => 3a + 3ib - 2a + 2ib = -30 => a + 5ib =...

Latest answer posted May 7, 2011, 9:38 pm (UTC)

algebra1

We have 5x + 4y = 9 and 3x + 2y = 5. We need to find 4x + 3y. Here we don't need to find x and y. The required result can be obtained by adding 5x + 4y = 9 and 3x + 2y = 5 => 5x + 4y + 3x + 2y =...

Latest answer posted March 3, 2011, 12:39 am (UTC)

algebra1

We have to solve -2x - y = -9 ...(1) 5x - 2y = 18 ...(2) 2*(1) - (2) => -4x - 2y - 5x + 2y = -18 - 18 => -9x = -36 => x = 4 substitute in (1) -2x - y = -9 => y = -2*4 + 9 => y = 1...

Latest answer posted May 7, 2011, 11:22 am (UTC)

algebra1

We have to determine a and b given that for x*y = xy + 2ax + by, the * operator is commutative. x*y = y*x => xy + 2ax + by = yx + 2ay + bx => 2ax + by = 2ay + bx equating the coefficients of...

Latest answer posted February 22, 2011, 1:27 am (UTC)