algebra1 Questions and Answers

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

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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

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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

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algebra1

If an operation % is commutative it implies that x%y = y%x. As * is commutative for x*y = xy + 4mx + 2ny, we have: xy + 4mx + 2ny = yx + 4my + 2nx => 4mx + 2ny = 4my + 2nx => x(2m - n) + y(n...

Latest answer posted March 14, 2011 5:25 pm UTC

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algebra1

It is given that log(72) 48 = a and log(6) 24 = b a = log(72) 48 = log(6) 48/ log(6) 72 => log(6) (6*8)/log(6) 6*12 => [1 + log(6) 8]/[1 + log(6) 12] => [1 + 3*log(6) 2]/[2 + log(6) 2] b...

Latest answer posted May 7, 2011 6:59 am UTC

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algebra1

The triangle has vertexes A(2,3), B(1, -6), C(0,3) AB = sqrt( 1 + 81) = sqrt 82 BC = sqrt(1 + 81) = sqrt 82 AC = 2 The area of a triangle is given by the Heron's formula as sqrt [s(s - a)(s - b)(s...

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

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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

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algebra1

The solutions of the quadratic equation y^2 + 6y + 9 = 0 can easily be found without finding the discriminant. y^2 + 6y + 9 = 0 => y^2 + 3y + 3y + 9 = 0 => y( y + 3) + 3(y + 3) = 0 => (y +...

Latest answer posted March 7, 2011 9:59 pm UTC

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algebra1

The square root of x^2 - 5x + 4 is a real value if x^2 - 5x + 4 >=0 x^2 - 5x + 4 = 0 => x^2 - 4x - x + 4 =0 => x(x - 4) - 1(x - 4) = 0 => (x - 1)(x - 4) = 0 The roots are x = 1 and x =...

Latest answer posted January 26, 2011 12:39 am UTC

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algebra1

We have to find the value of x for which log(x+1) (x^2-3x+2) = log(x+1) (x^2-2x) As the base of the log is the same, we can equate x^2 - 3x + 2 = x^2 - 2x => x = 2 But for x = 2, we get log 0...

Latest answer posted March 8, 2011 1:58 am UTC

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algebra1

We have to solve for x given that x^(ln x / ln 2) = 6 - 8/x^(ln x / ln 2). x^(ln x / ln 2) = 6 - 8/x^(ln x / ln 2) Let x^(ln x / ln 2) = y => y = 6 - 8/y => y^2 - 6y + 8 = 0 => y^2 - 4y -...

Latest answer posted March 21, 2011 3:40 am UTC

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algebra1

We have to solve 13/20-7/10x=0.5 for x 13/20-7/10x=0.5 => 13x/20x - 14/20x = 1/2 => (13x - 14)/20x = 1/2 => 13x - 14 = 10x => 3x = 14 => x = 14/3 The value of x = 14/3

Latest answer posted May 11, 2011 5:53 am UTC

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algebra1

We have to solve f(x) > -1 for f(x)=(x-1)(x+1)/(x+2)(x-2) f(x) > -1 => (x-1)(x+1)/(x+2)(x-2) > -1 => (x-1)(x+1) > -1*(x+2)(x-2) => x^2 - 1 > -1* (x^2 - 4) => x^2 - 1 >...

Latest answer posted January 21, 2011 3:26 am UTC

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algebra1

4^(2x - 5) = 64 => 4^(2x - 5) = 4^3 as 4 is equal we equate the exponent 2x - 5 = 3 => 2x = 3 + 5 => x = 8/2 => x = 4 Now we have the 6 terms of which 4 is one term, so the probability...

Latest answer posted January 26, 2011 12:50 am UTC

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algebra1

We have to find the point of intersection of y = 1008 - 2x and y = -x Substitute y = -x in y = 1008 - 2x => -x = 1008 - 2x => 2x - x = 1008 => x = 1008 Therefore the point of intersection...

Latest answer posted January 26, 2011 12:58 am UTC

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algebra1

We have to simplify: 4t^2- 16/8/t - 2/6 4t^2- 16/8/t - 2/6 => 4t^2- (16/8)/t - (2/6) 16/8 = 2 and 2/6 = 1/3 => 4t^2- 2/t - 1/3 We can simplify 4t^2- 16/8/t - 2/6 as 4t^2- 2/t - 1/3.

Latest answer posted May 11, 2011 6:32 am UTC

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algebra1

We are given that y = 3x / (x^2 - 9). We have to determine m and n if y = m/(x - 3) + n/(x +3) Equate the two expressions for y. 3x / (x^2 - 9) = m/(x - 3) + n/(x +3) => 3x / (x^2 - 9) = [m(x +...

Latest answer posted January 26, 2011 1:16 am UTC

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algebra1

An increasing linear function is of the form y = ax + b where a is positive. It is given that f(f(x))=4x+3 => f(ax + b) = 4x + 3 => a(ax + b) + b = 4x + 3 => a^2x + ab + b = 4x + 3 =>...

Latest answer posted May 7, 2011 11:39 am UTC

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algebra1

We have to solve: (0.25)^(x-4)=<(1/16)^x (0.25)^(x-4)=<(1/16)^x => (1/4)^(x- 4) =< (1/16)^x => (1/4)^(x- 4) =< (1/4)^2x As (1/4) is less than 1, we have 1/ 4 > (1/4^2)...

Latest answer posted January 29, 2011 1:57 am UTC

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algebra1

We have the points P(7,11) and Q(-2,4), and we need to find the length, slope and the midpoint of the line segment joining them. The length of the line segment is: sqrt ((7 + 2)^2 + (11 - 4)^2)...

Latest answer posted May 11, 2011 6:39 am UTC

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algebra1

The function f(x) = x^2 - tx - 3. The point (2, 9) lies on the curve. => 9 = 2^2 - t*2 - 3 => 9 = 4 - 2t - 3 => 8 = 2t => t = 4 The coefficient t = 4

Latest answer posted May 11, 2011 6:41 am UTC

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algebra1

We need to find x if 7^(3x-5) = 7^-2 7^(3x-5) = 7^-2 as the exponent 7 is common we can equate the exponent => 3x - 5 = -2 => 3x = 5 - 2 => 3x = 3 => x = 3/3 => x = 1 The solution...

Latest answer posted April 20, 2011 8:54 am UTC

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algebra1

We have to prove that a^2 - 4a + b^2 + 10b + 29>=0, for real values of a and b. a^2 - 4a + b^2 + 10b + 29 => a^2 - 4a + 4 + b^2 + 10b + 25 => (a - 2)^2 + (b + 5)^2 The sum of squares of...

Latest answer posted June 12, 2011 6:56 pm UTC

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algebra1

For solving a general inequality of the form ax + b > c, we isolate the variable x to one side. ax + b > c => ax > c - b => x > (c - b)/a The values of x that satisfy the...

Latest answer posted April 21, 2011 11:52 am UTC

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algebra1

y-3=3(x+1)If we are going to express this in slope-intercept form y=mx+b, then we have todistribute 3 to x+1.y-3=3x+3And, isolate the y by adding 3 on both sides of the...

Latest answer posted March 28, 2013 2:56 am UTC

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algebra1

Two graphs have a point of intersection if the equations of the graph have a solution. The equations of the graphs given are: f(x)=2x+1 and g(x)=x^2+x+1 Equating the two we get 2x + 1 = x^2 + x +...

Latest answer posted May 6, 2011 11:33 pm UTC

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algebra1

The terms a, b, c and d form a geometric progression. So we can write the terms as b = ar , c = ar^2 and d = ar^3 We have to prove (a - d)^2 = (b - c)^2 + (c - a)^2 + (d - b)^2 We start with the...

Latest answer posted February 19, 2011 11:29 pm UTC

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algebra1

The value of x if 4^(4x - 15) - 4 = 0 can be found by rewriting the equation as 4^(4x - 15) = 4 Now, as the base is the same on both the sides, equate the exponent 4x - 15 = 1 => 4x = 16 => x...

Latest answer posted May 5, 2011 12:12 am UTC

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algebra1

We have to simplify [(a+i)^3-(a-i)^3]/[(a+i)^2-(a-i)^2]. [(a+i)^3-(a-i)^3]/[(a+i)^2-(a-i)^2] we open the brackets => [a^3 + 3a^2*i+ 3a*i^2+ i^3 - a^3 + 3*a^2*i - 3*a*i^2 + i^3]/[(a^2 + 2ai + i^2...

Latest answer posted February 18, 2011 7:28 pm UTC

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algebra1

We have to solve the following set of simultaneous equations: x + 3y = 2 ...(1) 2x + 8y = 6 ...(2) 2*(1) - (2) => 2x + 6y - 2x - 8y = 4 - 6 => -2y = -2 => y = -2/-2 => y = 1 substitute...

Latest answer posted February 12, 2011 10:10 pm UTC

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algebra1

We have to solve : C(n+2,4) = n^2-1 We know that C(n,k) = n!/k!(n-k)! C(n+2,4) = n^2-1 => (n +2)! / 4! * ( n - 2)! = n^2 - 1 => n*(n -1)(n + 1)((n +2) / 4! = ( n - 1)( n + 1) => n(n + 2) =...

Latest answer posted February 20, 2011 12:24 am UTC

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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

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algebra1

The system of equations we have is: (x - y)^2 - 2xy = 6 ...(1) x^2 + y^2 = 2 ...(2) (1) => (x - y)^2 - 2xy = 6 => x^2 + y^2 - 2xy - 2xy = 6 From (2) x^2 + y^2 = 2 => -4xy + 2 = 6 => xy...

Latest answer posted May 1, 2011 1:13 am UTC

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algebra1

We are given that f(x) = x^2 - 4 and g(x) = x + 3. We have to solve for x given that f(g(x)) = 0. f(g(x)) = 0 => f( x +3) = 0 => (x +3)^2 - 4 = 0 => (x + 3 - 2)(x + 3 + 2) = 0 (x + 3 - 2)...

Latest answer posted February 26, 2011 1:54 am UTC

1 educator answer

algebra1

We need to find (x^2 + 1) - (x^2 - 4x + 1)^2 (x^2 + 1) - (x^2 - 4x + 1)^2 => x^2 + 1 - [(x^2 - 4x) + 1]^2 => x^2 + 1 - (x^2 - 4x)^2 - 1^2 - 2(x^2 - 4x) => x^2 + 1 - x^4 - 16x^2 + 8x^3 - 1...

Latest answer posted March 3, 2011 1:41 am UTC

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algebra1

We have to solve : log(4) (x+4) + log(4) 1/(x+1) - 1 = 0 Use the relation log a + log b = log (a*b) log(4) (x+4) + log(4) 1/(x+1) - 1 = 0 => log(4) (x + 4) / (x + 1) - 1 = 0 => log(4) (x +...

Latest answer posted February 13, 2011 12:38 am UTC

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algebra1

We have to simplify : [1/(x-1)-1/(x+1)+1]*(x+1)/(x^2+1) [1/(x-1)-1/(x+1)+1]*(x+1)/(x^2+1) => [((x+1) - (x - 1) + (x^2 - 1))/(x^2 -1)]*(x+1)/(x^2+1) => [(x + 1 - x + 1 + x^2 - 1)/(x^2...

Latest answer posted March 3, 2011 2:00 am UTC

1 educator answer

algebra1

We have to solve 3^(3x-9) = 1/81 for x 3^(3x-9) = 1/81 => 3^(3x-9) = 3^(-4) as the base is the same equate the exponent 3x - 9 = -4 => 3x = 5 => x = 5/3 The solution is 5/3

Latest answer posted May 3, 2011 1:07 pm UTC

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algebra1

We have to solve 11^(t+1) -1=4*11^t for t. 11^(t+1) -1=4*11^t => 11^t*11 - 1 = 4*11^t let 11^t = x => 11x - 1 = 4x => 7x = 1 => x= 1/7 11^t = 1/7 Here the bases are different, so we...

Latest answer posted February 17, 2011 5:31 pm UTC

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algebra1

We have to solve 5^(2x+1)=5^-x^2 5^(2x+1)=5^-x^2 As the base is the same we can equate the exponent: 2x + 1 = -x^2 => x^2 + 2x + 1 = 0 => (x + 1)^2 = 0 => x = -1 The required solution is x...

Latest answer posted February 21, 2011 1:19 am UTC

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algebra1

The equation to be solved is : (3x+7)(x-1) = 24 (3x+7)(x-1) = 24 => 3x^2 + 4x - 7 = 24 => 3x^2 + 4x - 31 = 0 x1 = -4/6 + sqrt (16 + 372) /6 => -2/3 + (sqrt 97)/6 x2 = -2/3 - (sqrt 97)/6...

Latest answer posted May 3, 2011 11:36 am UTC

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algebra1

We have to solve for x given that lg x + 15*lg 15 = 0 lg x + 15*lg 15 = 0 use the rule a*lg b = lg ( b^a) => lg x + lg 15^15 = 0 => lg x = - lg 15^15 => lg x = lg ( 1/15^15)) => x = 1/...

Latest answer posted February 20, 2011 1:56 am UTC

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algebra1

The expressions 64x^2 + a + 121y^2 and 25x^4/16- b + 16x^2/25 are squares. (8x + 11y)^2 is a square and we can equate this to 64x^2 + a + 121y^2 => (8x + 11y)^2 = 64x^2 + a + 121y^2 => 64x^2...

Latest answer posted April 5, 2011 12:22 am UTC

1 educator answer

algebra1

It is given that |3x+6|=9 As the absolute value of 3x + 6 is equal to 9 either 3x + 6 = 9 or -(3x + 6) = 9 3x + 6 = 9 => 3x = 3 => x = 1 -(3x + 6) = 9 => 3x + 6 = -9 => 3x = -15 => x...

Latest answer posted May 7, 2011 10:51 am UTC

2 educator answers

algebra1

The equation of the parabola is y = x^2 - 6x + 8 y = x^2 - 6x + 8 => x^2 - 6x + 9 - 1 => (x - 3)^2 -1 This is the standard form of a parabola y = (x - h)^2 + k The vertex is at ( 3, -1)

Latest answer posted April 18, 2011 12:15 am UTC

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algebra1

We have to solve the equation: log(x) 2 + log(2x) 2 = log(4x) 2 Now log(a) b = 1/ log (b) a So we can write log(x) 2 + log(2x) 2 = log(4x) 2 => 1/log (2) x + 1/ log 2( 2x) = 1/ log 2 ( 4x) =>...

Latest answer posted January 20, 2011 4:37 pm UTC

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algebra1

We have three equations to solve for z. x + 2y + 3z =1 ...(1) -x - y + 3z = 2 ...(2) -6x + y + z = -2 ...(3) x + 2y + 3z =1 => x = 1 - 2y - 3z ...(4) substitute in (2) -x - y + 3z = 2 => -1 +...

Latest answer posted January 23, 2011 7:51 pm UTC

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algebra1

Here we are given (x-9)^1/4 - (x+2)^1/3 = -1 and we have to find x! As factorial is defined only for an integer, the fourth root of x - 9 should be 1 less than the third root of x + 2 We have: 1^4...

Latest answer posted February 10, 2011 5:39 pm UTC

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algebra1

We have to solve x^8 - 64 = 0 Now 64 = 2^6 x^8 - 64 = 0 => x^8 = 64 => x^8 = 2^6 => x = 2^(6/8) => x = 2^(3/4) => x = 8^(1/4) => x= sqrt [2*sqrt 2] and x = -sqrt [2*sqrt 2] The...

Latest answer posted February 10, 2011 5:42 pm UTC

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algebra1

We have to expand (2 + 3x)^5 Using the binomial theorem: (2 + 3x)^5 = 2^5 + 5*2^4*(3x) + 10*2^3*(3x)^2 + 10*2^2*(3x)^3 + 5*2^1*(3x)^4 + (3x)^5 => 32 + 240x + 720x^2 + 1080x^3 + 810x^4 + 243x^5...

Latest answer posted January 27, 2011 3:31 am UTC

1 educator answer

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