2.8.1
By definition of the derivative,
We have
and
Combine these fractions into one with a common denominator:
Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:
Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :
3.1.1.
![f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3](https://tex.z-dn.net/?f=f%28x%29%20%3D%204x%5E5%20-%20%5Cdfrac1%7B4x%5E2%7D%20%2B%20%5Csqrt%5B3%5D%7Bx%7D%20-%20%5Cpi%5E2%20%2B%2010e%5E3)
Differentiate one term at a time:
• power rule
![\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%27%20%3D%20%5Cleft%28x%5E%7B1%2F3%7D%5Cright%29%27%20%3D%20%5Cdfrac13%20x%5E%7B-2%2F3%7D%20%3D%20%5Cdfrac1%7B3x%5E%7B2%2F3%7D%7D)
The last two terms are constant, so their derivatives are both zero.
So you end up with
25 goes into 200 exactly 8 times.
Answer:
The correct answer is b.
Step-by-step explanation:
The wave equation is given generally as:
c(x, t) = Acos(kx - wt)
Where A = amplitude
k = wave number
w = angular frequency.
x = horizontal distance moves by the wave.
t = time
The options show to us that the wave depends only on t and not (x, t).
Hence, the wave equation becomes:
c(t) = Acos(wt)
Given that:
A = 5 V
f = 1 * 10⁶ Hz
Angular Frequency, w, is given as:
w = 2πf
w = 2 * π * 1 * 10⁶ Hz
w = 2π(1 * 10⁶)
The wave equation becomes:
c(t) = 5cos(2*π*1*10⁶)
The correct answer is b.
Use Ga_thMath(u) (brainly doesn't allow me to type it) To use the app u need to take a pic of the problem and then it will process it and you'll get ur answer ASAP(most of the time). Many questions have been asked before so search it on brainly.
