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Math
We wish to evaluate the expression `10^(((x+y)/2) - 3z)` First, simplify the power into seperate terms involving each of `x`, `y` and `z` only: `10^(((x+y)/2) - 3z) = 10^(x/2 + y/2 - 3z)` Then...
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Math
First evaluate the expression under the square root sign: `x^2 + 2x + (y-1)^2 - 2xy = x^2 + 2x + y^2 - 2y + 1 - 2xy` `= x^2 + y^2 -2xy + 2(x-y) + 1` Now `(x-y)^2 = x^2 + y^2 - 2xy` So that...
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Math
a) We want the roots of `x^3 +3x^2 -4x + d=0` If we are given that two of the roots are opposites then we have that `x^3 + 3x^2 - 4x + d = (x-a)(x+a)(x-b)` `= (x^2 -a^2)(x-b)` Multiply this out...
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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...
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algebra1
You'll solve this equation involving combinations using the factorial formula for combinations of n elements taken k at a time: C(n,k) = n!/k!(n - k)! Let's evaluate C(n+1 , 3) = (n+1)!/3!(n+1-3)!...
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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...
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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!...
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algebra1
We'll apply the quotient rule of logarithms: log (a/b) = log a - log b According to this rule, we'll get: log (2x-5)/(x^2+3) = log (2x-5) - log(x^2+3) We'll re-write the equation: log (2x-5) -...
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algebra1
We'll replace the result of the difference x - y by z, at the numerator of the 1st option. (x-y)/2z = z/2z We'll simplify and we'll get: (x-y)/2z = 1/2 Since the result is not equal 2, we'll reject...
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algebra1
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Multiply initial fraction (3x - 5) and the 2nd by (4x-1)[3x(3x - 5) + 2x(4x-1)]/(4x-1)(3x-5)Remove brackets:(9x^2 - 15x + 8x^2 - 2x)/(12x^2 - 20x - 3x + 5)(17x^2 - 17x)/(12x^2 - 23x + 5)Square the...
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algebra1
To determine the expression of f(x^2), we'll simply replace x by x^2 in the expression of the function: f(x^2) = 4*(x^2)^2 - 3 f(x^2) = 4x^4 - 3 The found expression of f(x^2) is f(x^2) = 4x^4 - 3
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algebra1
The linear function put in standard form is: f(x) = ax + b Since the graph of the function is passing through the points (1,2) and (3,1), that means that if we'll substitute the coordinates of the...
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algebra1
We'll apply quadratic formula to determine the roots: b1 = [-(-21)+sqrt((-21)^2 + 4*108)]/2*1 b1 = (21+sqrt9)/2 b1 = (21+3)/2 b1 = 12 b2 = (21-3)/2 b2 = 9 We can write the quadratic expression as a...
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algebra1
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We'll determine the inverse of g(x) in this way. Let g(x) = y y = -2/(x+1) Now, we'll find x with respect to y. F y(x+1) = -2 We'll remove the brackets and we'll get: yx + y = -2 We'll isolate x to...
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algebra1
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We'll choose to re-write f(x): ( 2x^3 + 2x - 1 )/ (x^2 + 1) = (2x^3 + 2x)/(x^2 + 1) - 1/(x^2 + 1) We'll factorize by 2x the first fraction: (2x^3 + 2x)/(x^2 + 1) = 2x(x^2 + 1)/(x^2 + 1) We'll...
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algebra1
We'll write the rectangular form of any complex number is z = x + y*i. The trigonometric form of a complex number is: z = |z|(cos a + i*sin a) |z| = sqrt(x^2 + y^2) cos a = x/|z| sin a = y/|z|...
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algebra1
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We notice that 6^x = (2*3)^x But (2*3)^x = 2^x*3^x We'll subtract 5*2^x*3^x both sides: 3*2^2x + 2*3^2x - 5*2^x*3^x = 0 We'll divide by 3^2x: 3*(2/3)^2x - 5*(2/3)^x + 2 = 0 We'll note (2/3)^x = t...
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algebra1
We'll use the theorem of arithmetic mean to determine the terms of the arithmetic progression. 4 = (x+y)/2 => 8 = x+y (1) y = (4+12)/2 y = 16/2 y = 8 We'll substitute y into (1): 8 = x+8 We'll...
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algebra1
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Before solving the equation, we'll impose conditions of existence of the square root. 5x-6 >= 0 We'll subtract 6 both sides: 5x >= 6 We'll divide by 5: x >=6/5 The interval of admissible...
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algebra1
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We'll denote the point with that has equal coordinates as M(m,m). Since the point is located on the line y = 0.5x - 0.5, it's coordinates verify the expression of the line. We'll put y = f(x) and...
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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...
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algebra1
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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
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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)...
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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.
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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
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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 -...
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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...
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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 =...
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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 =...
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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 =...
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algebra1
We'll write the given expresison as a fraction, using the negative power property: (7x-x^2)^-1 = 1/(7x-x^2) We'll get 2 elementary fractions because we notice 2 factors at denominator. 1/(7x-x^2)...
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algebra1
We'll remove the brackets, using FOIL method: (-2i+5)(i+7) = -2i^2 - 14i + 5i + 35 We know that i^2 = -1 (-2i+5)(i+7) = 2 - 9i + 35 We'll combine real parts: (-2i+5)(i+7) = 37 - 9i The result of...
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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)...
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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...
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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 =...
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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...
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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
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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 -...
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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 +...
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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...
