
MathWe 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...

MathFirst evaluate the expression under the square root sign: `x^2 + 2x + (y1)^2  2xy = x^2 + 2x + y^2  2y + 1  2xy` `= x^2 + y^2 2xy + 2(xy) + 1` Now `(xy)^2 = x^2 + y^2  2xy` So that...

Matha) 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 = (xa)(x+a)(xb)` `= (x^2 a^2)(xb)` Multiply this out...

algebra1y3=3(x+1)If we are going to express this in slopeintercept form y=mx+b, then we have todistribute 3 to x+1.y3=3x+3And, isolate the y by adding 3 on both sides of the...

algebra1Solve the following equation for n C (n+1, 3) = C(n,2) I know the answer is 5  since I did trial...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+13)!...

algebra1We 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...

algebra1k! = 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!...

algebra1We'll apply the quotient rule of logarithms: log (a/b) = log a  log b According to this rule, we'll get: log (2x5)/(x^2+3) = log (2x5)  log(x^2+3) We'll rewrite the equation: log (2x5) ...

algebra1We'll replace the result of the difference x  y by z, at the numerator of the 1st option. (xy)/2z = z/2z We'll simplify and we'll get: (xy)/2z = 1/2 Since the result is not equal 2, we'll reject...

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algebra1Multiply initial fraction (3x  5) and the 2nd by (4x1)[3x(3x  5) + 2x(4x1)]/(4x1)(3x5)Remove brackets:(9x^2  15x + 8x^2  2x)/(12x^2  20x  3x + 5)(17x^2  17x)/(12x^2  23x + 5)Square the...

algebra1To 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

algebra1The 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...

algebra1We'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 = (213)/2 b2 = 9 We can write the quadratic expression as a...

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algebra1We'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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algebra1We'll choose to rewrite 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...

algebra1We'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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algebra1We 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...

algebra1We'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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algebra1Before solving the equation, we'll impose conditions of existence of the square root. 5x6 >= 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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algebra1We'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...

algebra1The 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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algebra1determine coefficients of quadraticGiven (2,9) a point on the graph of f(x), what is t, if f(x) =...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

algebra1We 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)...

algebra1We 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.

algebra1We have to solve 13/207/10x=0.5 for x 13/207/10x=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

algebra1To 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 ...

algebra1A 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...

algebra1Consecutive 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 =...

algebra1We 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 =...

algebra1We 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 =...

algebra1We'll write the given expresison as a fraction, using the negative power property: (7xx^2)^1 = 1/(7xx^2) We'll get 2 elementary fractions because we notice 2 factors at denominator. 1/(7xx^2)...

algebra1We'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...

algebra1It 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)...

algebra1It 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...

algebra1It 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 =...

algebra1The equation to be solved is : (3x+7)(x1) = 24 (3x+7)(x1) = 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...

algebra1We have to solve 3^(3x9) = 1/81 for x 3^(3x9) = 1/81 => 3^(3x9) = 3^(4) as the base is the same equate the exponent 3x  9 = 4 => 3x = 5 => x = 5/3 The solution is 5/3

algebra1The equation to be solved is (x4)^1/2=1/(x4) (x4)^1/2=1/(x4) 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 ...

algebra1We 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 +...

algebra1The 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...
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