A function is differentiable if you can find the derivative at every point in its domain. In the case of f(x) = |x+2|, the function wouldn't be considered differentiable unless you specified a certain sub-interval such as (5,9) that doesn't include x = -2. Without clarifying the interval, the entire function overall is not differentiable even if there's only one point at issue here (because again we look at the entire domain). Though to be fair, you could easily say "the function f(x) = |x+2| is differentiable everywhere but x = -2" and would be correct. So it just depends on your wording really.
Answer:
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Step-by-step explanation:
Since the Capital City's population is of Medfield's is 7000 please tell if you forgot something. Good luck
Answer:
The solution to the differential equation
y' = 1 + y²
is
y = tan x
Step-by-step explanation:
Given the differential equation
y' = 1 + y²
This can be written as
dy/dx = 1 + y²
Separate the variables
dy/(1 + y²) = dx
Integrate both sides
tan^(-1)y = x + c
y = tan(x+c)
Using the initial condition
y(0) = 0
0 = tan(0 + c)
tan c = 0
c = tan^(-1) 0 = 0
y = tan x
Answer:
Step-by-step explanation:
![\sin \: 30 \degree = \frac{ \sqrt{5} }{x} \\ \\ \frac{1}{2} = \frac{ \sqrt{5} }{x} \\ \\x = 2 \sqrt{5}](https://tex.z-dn.net/?f=%20%5Csin%20%5C%3A%2030%20%5Cdegree%20%3D%20%20%5Cfrac%7B%20%5Csqrt%7B5%7D%20%7D%7Bx%7D%20%20%5C%5C%20%20%5C%5C%20%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20%20%5Cfrac%7B%20%5Csqrt%7B5%7D%20%7D%7Bx%7D%20%20%5C%5C%20%20%5C%5Cx%20%3D%202%20%5Csqrt%7B5%7D%20)