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What is the derivative of fx where fx equals 2x-3 all under square root?
f(x) = √(2x -3) f(x) = (2x - 3)^(1/2) f'(x) = (1/2)[(2x - 3)^(1/2 - 1)](2) f'(x) = (2x - 3)^(-1/2) f'(x) = 1/[(2x - 3)^(1/2)] f'(x) = 1/√(2x -3)
X2 over 3x2 -5x-2 -2x over 3 x plus 1 x 1 over x-2?
x2/(3x2 - 5x - 2) - [2x/ (3x + 1)][1/(x - 2)] = x2/(3x2 - 6x + x - 2) - 2x/(3x + 1)(x - 2) = x2/[3x(x - 2) + (x - 2)] - 2x/(3x + 1)(x - 2) = x2/(3x + 1)(x - 2) - 2x/(3x + 1)(x - 2) = (x2 - 2x)/(3x + 1)(x - 2) = x(x - 2)/(3x + 1)(x - 2) = x/(3x + 1)
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
What are the solutions to the simultaneous equations of x squared plus xy plus y squared equals 7 and 2x plus y equals 1 showing work?
If: 2x+y = 1 Then: y = 1-2x If: x^2 + xy + y^2 = 7 Then: x^2 +x(1-2x) +(1-2x)^2 = 7 So: x^2 +x-2x^2 +1-4x+4x^2 -7 = 0 Collecting like terms: 3x^2 -3x -6 = 0 Divide all terms by 3: x^2 -x -2 = 0 Factorizing (x+1)(x-2) = 0 => x = -1 or x = 2 Solutions by substitution: (-1, 3) and (2, -3)
What is (1 plus cos x)(1- cos x)?
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
What is the sum of 2x plus 1 and x times x minus 2x plus one?
(2x + 1) + (x*x - 2x + 1) = x^2 + 2x - 2x + 1 + 1 = x^2 + 2
What is the derivative of fx where fx equals 2x-3 all under square root?
f(x) = √(2x -3) f(x) = (2x - 3)^(1/2) f'(x) = (1/2)[(2x - 3)^(1/2 - 1)](2) f'(x) = (2x - 3)^(-1/2) f'(x) = 1/[(2x - 3)^(1/2)] f'(x) = 1/√(2x -3)
What is the solution of the equation 2x - 2 ( x - 1 ) = 2x + 1?
x = 1/2
What is the answer to 3 equals 2x plus 1?
x = 1 ------------- 3 = 2x + 1 -1 -1 2=2x /2 /2 1=x
X2 over 3x2 -5x-2 -2x over 3 x plus 1 x 1 over x-2?
x2/(3x2 - 5x - 2) - [2x/ (3x + 1)][1/(x - 2)] = x2/(3x2 - 6x + x - 2) - 2x/(3x + 1)(x - 2) = x2/[3x(x - 2) + (x - 2)] - 2x/(3x + 1)(x - 2) = x2/(3x + 1)(x - 2) - 2x/(3x + 1)(x - 2) = (x2 - 2x)/(3x + 1)(x - 2) = x(x - 2)/(3x + 1)(x - 2) = x/(3x + 1)
2x cubed - 2x squared - x plus 1?
(x - 1)(2x^2-1)
What is -x-2 equals 1-2x?
x=3 -x-2=1-2x Add 2x to either side x-2=1 Add 2 to either side x=3
How do you work out x plus 2 equals 2x-1?
x+2=2x-1-x+1 -x+1-------------3=x
What is 2 X cubed minus 2 X equals 0?
2x^3 - 2x = 0 2x(x - 1)(x + 1) = 0 x = 0, 1, -1
2x2 5x -2?
=2x2+x+4x+2=x(2x+1)+2(2x+1)=(2x+1)(x+2) by spaz licker enjoy slag ;)
What is the integral of cos raised to 2 times x?
∫cos2(x).dxUse the identity cos2(x) = (1/2)(1+cos(2x))∫(1/2)(1+cos(2x))dxPull out constant:(1/2)∫(1+cos(2x))dxIntegrate:(1/2)(x + sin(2x)/2) + CSimplify:x/2 + sin(2x)/4 + CThe identity sin(2x) = 2sin(x)cos(x) can be used to rewrite it as(x + sin(x)cos(x))/2 + C
How to simplify sin x2 cos x2?
If you mean: sin2(x) cos2(x) then it can be simplified by noting that the square of the sine of x is equal to (1 - cos(2x)) ÷ 2 and the square of the cosine of x is equal to (1 + cos(2x)) ÷ 2. We can then simplify further: sin(x)2cos(x)2 = [(1 - cos(2x)) / 2][(1 + cos(2x)) / 2] = (1 - cos(2x))(1 + cos(2x)) / 2 = (1 - cos2(2x)) / 2 Also note that 1 - cos2(x) = sin2(x), so we can then say: = sin2(2x) / 2
