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What is the derivative of 2 to the power of x?
if f(x)=kx, f'(x)=ln(k)*kx. Therefore, the derivative of 2x is ln(2)*2x.
What is the factor of k plus 5k?
k + 5k = 6k 2 x 3 x k
how do i solve 4x-9=k for x?
4x-9=k 4x=k+9 Divide by 4 to isolate the x x=(1/4)k+(9/4) x=1/4 k+ 9/4
Fx equals x cubed plus 4?
x = ¼x^4 + 4x + k where k can be any number
What is x to the fifth power times x to the fourth?
x to the ninth power OR x to the 9th power OR x^9
If K to the power of one third times K to the power of one third times K to the power of one third equals 1 why is x equivalent to x to the k power?
racist
How do you generate a moment generating function of a poisson distribution?
Let X have a Poisson distribution.Therefore Pr(X=x) = e^(-k) * k^x/x! where x = 0, 1, 2, ...Normally the parameter is the Greek letter, lambda, but this site does not allow any such symbols!Then the moment generating function of the Poisson distribution isE[exp(tx)] = Sum[e^(tx)*Pr(X=x)] where the sum is over all non-negative integers, x.= Sum[e^tx * e^(-k) * k^x/x!]= Sum[(e^t)^x * e^(-k) * k^x/x!]= e^(-k) * Sum[(e^t)^x * k^x/x!]= e^(-k) * Sum[(k*e^t)^x /x!]= e^(-k) * e^(k*e^t)= e^[k*(e^t - 1)]
How do you find the LCM of k to the 2nd power k to the 2nd power-1 and k to the 2nd power minus 2k plus 1?
Factor them. k2 = k x k k2 - 1 = (k - 1)(k + 1) k2 - 2k + 1 = (k - 1)(k - 1) Combine the factors, eliminating duplicates. k2(k + 1)(k - 1)(k - 1) = k5 - k4 - k3 + k2, the LCM
What is k equals xy for x?
k = x y x = k/y
What is the derivative of 2 to the power of x?
if f(x)=kx, f'(x)=ln(k)*kx. Therefore, the derivative of 2x is ln(2)*2x.
What k -k equals 2?
k and -k right? -k x -1 =k k+k= 2 k= 1 unless you mean multiply then that would be -k x-1 =k k x k= 2 1.4142 rounded to the nearest ten thousandth
What is the amperage for a 480 volt 108 K w unit?
Use the formula: power = voltage x current x power factor If you don't know the power factor, assume it is 1 - this is close enough for many devices. In units: power (in watts) = voltage (in volts) x current (in amperes)
How do you find the value of k when the area enclosed by x-axis and the curve y equals root x and the line x equals k is 18?
The area under a curve (ie between the curve and the x-axis) is found by integrating the curve with respect to x: A = ∫ y dx When limits to the integration are given, then the value of the area is the difference between the value of the integral at the limits. The lower limit is where the curve y = √x meets the x-axis, ie at y = 0 → √x = 0 → x = 0 The upper limit is the line x = k → the area is given by ∫ √x dx between 0 and k To integrate √x write it in power format: √x = x^½, giving: ∫ √x dx = ∫ x^½ dx = ⅔x^(3/2) + c The difference between the limits is the area: (⅔k^(3/2) + c) - (⅔0^(3/2) + c) = 18 → ⅔k^(3/2) = 18 → k^(3/2) = 18 ÷ ⅔ = 27 → k = 27^(2/3) = (27^⅓)² = 9 Thus k = 9.
What is the integral of 10x?
∫ 10x dx Factor out the constant: 10 ∫ x dx Therefore, by the power rule, we obtain: 10x(1 + 1)/(1 + 1) + k = 10x²/2 + k = 5x² + k
What is the total power loss if 2 k and 1 k parallel-connected resistors have an IT of 3 mA?
I'm assuming that by ' IT ' you mean ' It ' for 'total current'.-- The effective resistance of 2k and 1k in parallel is 2/3 k ohms.-- The power dissipation is I2R = 0.003 x 2/3 k = 2 watts
Pentax k-x or canon t1i?
K-x. In red!
What is the factor of k plus 5k?
k + 5k = 6k 2 x 3 x k
