Questions and Answers for algebra1

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

2 educator answers

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

2 educator answers

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

2 educator answers

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

1 educator answer

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

2 educator answers

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

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

1 educator answer

algebra1

Given the equation: 1- sqrt(13+3x^2) = 2x We need to solve for x. First we will move 1 to the right side. ==> - sqrt(13+3x^2) = 2x -1 Now we will square both sides. ==> (13+3x^2) = (2x-1)^2...

Latest answer posted February 4, 2011 3:31 am UTC

1 educator answer

algebra1

We have to solve the equations: y^2 = x^2 - 9 ...(1) y = x - 1 ...(2) From (2) we get y = x - 1 => y^2 = (x - 1)^2 substitute in (1) (x - 1)^2 = x^2 - 9 => x^2 + 1 - 2x = x^2 - 9 => -2x =...

Latest answer posted March 21, 2011 3:46 am UTC

1 educator answer

algebra1

The equation of the line passing through the points (x1, y1) and (x2, y2) is given by ( y - y1) = [( y2 - y1)/(x2 - x1)]( x - x1) As the line passes through ( -1 , 3) and (3 , 1) we can substitute...

Latest answer posted February 17, 2011 5:37 pm UTC

1 educator answer

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

1 educator answer

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

1 educator answer

algebra1

As z is a complex number z = x+ iy z' = x - iy z + z' = x + iy + x - iy = 2x z'*z^2 + i*z*z' = (x - iy)*(x +iy)^2 + i*(x - iy)(x + iy) => (x^2 + y^2)(x + iy) + i(x^2 + y^2) => x^3 + x^2yi +...

Latest answer posted January 24, 2011 12:39 am UTC

1 educator answer

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

1 educator answer

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

1 educator answer

algebra1

We have to solve: x + y - 14 = 0 ...(1) 4x - y - 11 = 0 ...(2) From (1) x + y - 14 = 0 => -y = x - 14 Substitute in (2) 4x + x - 14 - 11 = 0 => 5x - 25 = 0 => x = 25/5 => x = 5 y = 14 -...

Latest answer posted February 25, 2011 5:06 pm UTC

1 educator answer

algebra1

The equation to be solved is (x-4)^1/2=1/(x-4) (x-4)^1/2=1/(x-4) square both the sides => x - 4 = 1 / (x - 4)^2 => (x - 4)^3 = 1 => 1 - (x - 4)^3 = 0 => (1 - (x - 4))(1 + x - 4 + (x -...

Latest answer posted May 3, 2011 4:57 pm UTC

1 educator answer

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

1 educator answer

algebra1

We have k = (x^2 - 4)/( 2x - 5) k = (x^2 - 4)/( 2x - 5) => x^2 - 4 = 2kx - 5k => x^2 - 2kx + 5k - 4 = 0 As the roots of the quadratic equation are equal, b^2 - 4ac = 0 => (-2k)^2 - 4*( 5k...

Latest answer posted February 17, 2011 12:40 am UTC

1 educator answer

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

1 educator answer

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