algebra1 Questions and Answers

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 = 3x^2-12x substituting the values, the vertex is (12/6...

Latest answer posted May 7, 2011 11:34 am UTC

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algebra1

The partial fractions of (3x+2)/(x^2+x) have to be determined. (3x+2)/(x^2+x) => (3x+2)/x(x + 1) This can be written as A/x + B/(x + 1) A/x + B/(x + 1) = (3x+2)/[x(x + 1)] Ax + A + Bx = 3x + 2...

Latest answer posted April 30, 2011 9:21 pm UTC

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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

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algebra1

The equation of the linear function passing through the points (2,4) and (-4 , -2) is a straight line given by y + 2 = [ (4 + 2) / (2 + 4)]*( x + 4) => y + 2 = (6/6)*(x + 4) => y + 2 = x +...

Latest answer posted March 3, 2011 3:40 pm UTC

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algebra1

The equations to be solved are x = 53/11 - 16y/11...(1) x + 4y = 43 ...(2) (-4/11)*(2) + (1) => -4x/11 - 16y/11 + x = 53/11 - 16y/11 -4*43/11 => 7x/11 = -119/11 => x = -17 substitute in...

Latest answer posted April 30, 2011 8:47 pm UTC

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algebra1

We need to determine the solutions of u(v(t)) = 0 given that u(t) = t^2 - 16 and v(t) = t + 2. u(v(t)) = 0 => u( t + 2) = 0 => (t + 2)^2 - 16 = 0 => (t + 2)^2 - 4^2 = 0 => (t + 2 - 4)(t...

Latest answer posted March 3, 2011 3:49 pm UTC

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algebra1

The equation to be solved is 2x^4 - 11x^3 + 23x^2 - 19x + 5 = 0 We can see that for x = 1, 2 - 11 + 23 - 19 + 5 = 0 This gives us one solution as 1. 2x^4 - 11x^3 + 23x^2 - 19x + 5 = 0 => 2x^4 -...

Latest answer posted April 30, 2011 9:16 pm UTC

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algebra1

We have to factor a^4 - b^16 we use x^2 - y^2 = (x +y)(x - y) a^4 - b^16 => (a^2 - b^8)(a^2 + b^8) => (a - b^4)(a + b^4)(a^2 + b^8) Therefore a^4 - b^16 = (a - b^4)(a + b^4)(a^2 + b^8)

Latest answer posted January 22, 2011 1:38 am UTC

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algebra1

We have to solve the following by elimination 2x - y = 2 ...(1) 6x + 8y = 39 ...(2) 3*(1) - (2) => 6x - 3y - 6x - 8y = 6 - 39 => -11y = -33 => y = -33/-11 => y = 3 substituting y = 3 in...

Latest answer posted February 4, 2011 3:00 am UTC

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algebra1

The equation to be solved is (2x+48)^1/2 - x = 0 (2x+48)^1/2 - x = 0 => (2x+48)^1/2 = x square both the sides 2x + 48 = x^2 => x^2 - 2x - 48 = 0 => x^2 - 8x + 6x - 48 = 0 => x(x - 8) +...

Latest answer posted April 30, 2011 9:02 pm UTC

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algebra1

We have to solve (x+57)/6=x^2 (x+57)/6=x^2 => x + 57 = 6x^2 => 6x^2 - x - 57 = 0 The roots of a quadratic equation ax^2 + bx + c = 0 are given by [-b + sqrt (b^2 - 4ac)]/2a and [-b + sqrt...

Latest answer posted January 20, 2011 11:38 pm UTC

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algebra1

We have to prove that (x/(x+1)+1):(1-3x^2/(1-x^2))=(1-x)/(1-2x) Starting with the left hand side (x/(x+1)+1):(1-3x^2/(1-x^2)) => [(x/(x+1)+1)]/[(1-3x^2/(1-x^2))] =>...

Latest answer posted March 3, 2011 5:15 pm UTC

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algebra1

We have to solve for x given 64x^3+1331=0 64x^3 + 1331 = 0 => (4x)^3 + 11^3 = 0 => (4x + 11)( (4x)^2 - 4x*11 + 11^2) = 0 4x + 11 = 0 => x1 = -11/4 => x = -2.75 ( (4x)^2 - 4x*11 + 11^2)...

Latest answer posted January 22, 2011 2:11 am UTC

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algebra1

The roots of a quadratic equation, if irrational, are found as pairs. Here we have one roots as 1/3 + sqrt 2/3, the other root will be 1/3 - sqrt 2/3 We can determine the quadratic equation by...

Latest answer posted March 18, 2011 10:46 pm UTC

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algebra1

We have to find the common points of y=20-x and 3x-2y-6=0. This is equivalent to solving for x and y using the given equations. We have y=20-x. substitute this for y in 3x-2y-6=0 => 3x - 2( 20 -...

Latest answer posted January 22, 2011 2:31 am UTC

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algebra1

We have the line x = 42 - 14y. The other line is y = 2x - 11. Use x = 42 - 14y from the equation of the first line and substitute in the second y = 2x - 11 => y = 2( 42 - 14y) - 11 => y = 84...

Latest answer posted January 24, 2011 2:22 am UTC

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algebra1

We have the equation 11^(5x-6)=1/11^(-x-10) to solve for x. 11^(5x-6)=1/11^(-x-10) => 11^(5x - 6) = 11^[-( - x - 10)] => 11^(5x - 6) = 11^(x + 10) as the base is the same we can equate the...

Latest answer posted January 20, 2011 11:52 pm UTC

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algebra1

We can write 7.2 as 6*6/5 log 7.2 = log 6*6 / 5 = log 6^2 / 5 Now log (a^b) = b* log a and log (a/b) = log a - log b log (6^2 / 5) => 2 log 6 - log 5 as log 6 = m and log 5 = n => 2m - n...

Latest answer posted January 22, 2011 2:54 am UTC

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algebra1

We have to solve (x+5)^2-4(4x+5)=0 (x+5)^2 - 4(4x+5) = 0 open the brackets x^2 + 10x + 25 - 16x - 20 = 0 => x^2 - 6x + 5 = 0 => x^2 - 5x - x + 5 = 0 => x( x - 5) - 1(x - 5) = 0 => (x -...

Latest answer posted January 22, 2011 3:25 am UTC

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algebra1

We have to simplify [5/(4^-1-9^-1)]^1/2 [5/(4^-1-9^-1)]^1/2 => 5^(1/2)/(1/4 - 1/9)^(1/2) => 2*3*5^(1/2)/(9 - 4)^(1/2) => 2*3*5^(1/2)/5^(1/2) => 2*3 => 6 The result is 6.

Latest answer posted May 7, 2011 10:44 am UTC

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algebra1

We need to know the number of solutions of x - a = sqrt(x^2 - 1) if a = 0. x = sqrt(x^2 - 1) => x^2 = x^2 - 1 => x^2 - x^2 = 1 => 0 = 1 which is impossible. Therefore there are no solutions.

Latest answer posted May 7, 2011 10:46 am UTC

1 educator answer

algebra1

A function is odd if f(-x) = -f(x) and even if f(-x) = f(x) Here we have f(x) = y= 17x^3 - 12x^2 f(-x) = -17x^3 - 12x^2 f(x) = 17x^3 - 12x^2 -f(x) = -17x^3 + 12x^2 So we see that f(-x) = -17x^3 -...

Latest answer posted January 21, 2011 12:24 am UTC

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algebra1

The complex roots of a polynomial are always found as conjugate pairs. For the root 2 - i, the complex conjugate is 2 + i. The polynomial has the roots 2 - i and 2 + i Irrational roots are also...

Latest answer posted May 11, 2011 2:46 pm UTC

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algebra1

Complex numbers are of the form a + ib, with a real component a and an imaginary component which has a coefficient b. When you multiply two real numbers it is equivalent to multiplying 4 numbers....

Latest answer posted May 7, 2011 10:42 am UTC

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algebra1

We have to solve the system of equations: x^2+2y^2=10...(1) 3x^2-y^2=9 ...(2) (1) - 2*(2) => x^2 + 2y^2 - 3x^2 - 2y^2 = 10 - 18 => -2x^2 = -8 => x^2 = 4 Substitute in (2) 3*4 - y^2 = 9...

Latest answer posted February 22, 2011 3:04 am UTC

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algebra1

We have to solve the system of equations: 3x - 2y = -5 ...(1) -3x - y = -1 ...(2) (1) +(2) => 3x - 2y -3x - y = -5 - 1 => -3y = -6 => y = -6 / -2 => y = 2 Substitute in (1) 3x = -5 +...

Latest answer posted March 7, 2011 8:53 pm UTC

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algebra1

We have the polynomial P(x) = x^321+x^320+x^319+318 P(1) = 1^321 + 1^320 + 1^319 + 318 => 1 + 1 + 1 + 318 => 321 P(-1) = (-1)^321 + (-1)^320 + (-1)^319 + 318 => -1 + 1 - 1 + 318 => 317...

Latest answer posted March 7, 2011 8:56 pm UTC

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algebra1

We have to solve the equations 1/x=(y-1)/y and 3/x=(2y-1)/y 1/x=(y-1)/y => x = y/(y - 1) Substitute x = y/ (y - 1) in 3/x=(2y-1)/y => 3/x=(2y-1)/y => 3 /[y/ (y - 1)] = (2y-1)/y => 3(y -...

Latest answer posted January 21, 2011 1:32 am UTC

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algebra1

We can write a quadratic equation with roots a and b as: (x - a)(x - b) = 0 => x^2 - (a+b)x + ab = 0 as the product of the roots is -3. => x^2 - ( a + b) - 3 = 0 Now a and b can have several...

Latest answer posted January 26, 2011 4:55 pm UTC

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algebra1

It is not possible to solve the equation you have given t^4 + 10t - 11 = 0. Instead, here is the solution for t^4 + 10 t^2 - 11 = 0 let t^2 = x => x^2 + 10x - 11 = 0 => x^2 + 11x - x - 11 =...

Latest answer posted March 7, 2011 10:04 pm UTC

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algebra1

We have f(x) = 2x / 2(x^3 + x) Let y = f(x) = 2x / 2(x^3 + x) => y = 2x / 2x(x^2 +1) cancel 2x => y = 1/ x^2 + 1) => x^2 + 1 = 1/y => x^2 = 1/ y - 1 => x^ 2 = (1 - y)/y => x =...

Latest answer posted January 26, 2011 12:34 am UTC

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algebra1

We have the fraction 2i/(1 + i) Making the denominator a real number 2i/(1 + i) => 2i( 1 - i)/(1 + i)(1 - i) => (2i - 2i^2) / 1 - i^2 => (2i + 2)/ ( 1 + 1) => 1 + i Therefore we get (1...

Latest answer posted January 27, 2011 3:18 pm UTC

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algebra1

We have to solve the equation: 5^(t+1) = 4*5^t + 1 5^(t+1) = 4*5^t + 1 => 5^t * 5 = 4* 5^t + 1 let y = 5^t => y*5 = 4*y + 1 =>5y - 4y = 1 => y = 1 Now 5^t = y = 1 => t = 0 , as any...

Latest answer posted January 27, 2011 3:23 pm UTC

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algebra1

We have to simplify the fraction: 2/(2x*(6x+4)) and write it as a sum of simplified fractions. 2/(2x*(6x+4)) => 1/x*(6x + 4) This can be written as (A / x) + (B / (6x + 4)) => [A(6x + 4) +...

Latest answer posted April 13, 2011 10:29 pm UTC

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algebra1

The nth term of the binomial expansion of (5a+3b)^8 is given by C(8,n)*(5a)^(8-n)*(3b)^(8), and we have terms from n= 0 to n= 8. The middle terms for the expansion is C(8, 4)*(5a)^(8-4)*(3b)^(4)...

Latest answer posted January 29, 2011 12:53 am UTC

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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

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algebra1

We have to solve x^4 - 3x^2 + 2 = 0 x^4 - 3x^2 + 2 = 0 => x^4 - 2x^2 - x^2 + 2 = 0 => x^2(x^2 - 2) - 1(x^2 - 2) = 0 => (x^2 - 1)(x^2 - 2) = 0 x^2 = 1 => x = 1 , x = -1 x^2 = 2 => x =...

Latest answer posted May 11, 2011 5:02 am UTC

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algebra1

We have to find x given that 28^(4x-6)=1/784 28^(4x-6)=1/784 => 28^(4x-6)=1/28^2 => 28^(4x-6)=28^(-2) As the base is the same, we equate the exponent => 4x - 6 = -2 => 4x = 4 => x =...

Latest answer posted January 21, 2011 2:47 am UTC

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algebra1

We have to solve the equation log(2) [x(x-1)]=1 As the base of the logarithm is 2 we can rewrite it as: x(x-1) = 2^1 => x^2 - x = 2 => x^2 - x - 2 = 0 => x^2 - 2x +x - 2 = 0 => x(x - 2)...

Latest answer posted January 29, 2011 1:00 am UTC

1 educator answer

algebra1

We are given given that the ratio of the numerator and the denominator is 3/4 Let the numerator be N and the denominator be D N/ D = 3/4 => N = (3/4)*D (N - 8)/ ( D + 8) = 1/6 => 6N - 48 = D...

Latest answer posted February 14, 2011 9:43 pm UTC

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algebra1

We have to determine if 2x/(x+5) - x/(x-5) = 50/(25-x^2) has any solutions. 2x/(x+5) - x/(x-5) = 50/(25-x^2) => [2x(x - 5) - x(x + 5)] / (x^2 - 25) = 50 / (25-x^2) => 2x^2 - 10x - x^2 - 5x +...

Latest answer posted March 8, 2011 1:04 am UTC

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algebra1

We have to find all the real solution of (x-4)(x+8) < (x-8)(x+4). (x-4)(x+8) < (x-8)(x+4) => x^2 - 4x + 8x - 32 < x^2 - 8x + 4x - 32 => -4x + 8x < -8x + 4x => 4x < -4x...

Latest answer posted February 14, 2011 9:39 pm UTC

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algebra1

We have to solve for x: 2^( x^2 - 3x + 2) = 4^3 2^( x^2 - 3x + 2) = 4^3 => 2^( x^2 - 3x + 2) = 2^2^3 = 2^6 as the base 2 is equal we equate the exponent. x^2 - 3x + 2 = 6 => x^2 - 3x -4 = 0...

Latest answer posted January 26, 2011 12:46 am UTC

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algebra1

We have to solve sqrt(x^2 - 5) - sqrt(x^2 - 8) = 1 sqrt(x^2 - 5) - sqrt(x^2 - 8) = 1 square both the sides x^2 - 5 + x^2 - 8 - 2* sqrt [(x^2 - 5)(x^2 - 8)] = 1 => 2x^2 - 14 - 2* sqrt [(x^2 -...

Latest answer posted March 8, 2011 1:41 am UTC

1 educator answer

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

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algebra1

The linear function whose graph passes through (-4 , 0) and (1, 3) is the equation of the line passing through the two points. The equation of the line passing through (x1 , y1) and (x2, y2) is...

Latest answer posted January 26, 2011 12:43 am UTC

1 educator answer

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