A 1-dimensional interval (a, b) is continuous if for any k in (0, 1) the point a + k*(b-a) = a*(1-k) + k*b is also in the interval.
This is equivalent to the statement that every point on the line joining a and b is in the interval.
The above can be extended to more dimensions analogously.
A variable defined on a continuous interval as opposed to one that can take only discrete values.
Yes, it is a Continuous variable measured along an equidistant scale.
The linear discrete time interval is used in the interpretation of continuous time and discrete valued: Quantized signal.
Yes it is because it is a measurement of something usually entering into decimal figures and cannot be simply counted.
There are several different ways of defining continuity. The following is based on work done by Bolzano and Weierstrass.A function f(x), of a variable x is continuous at the point c if, given any positive number e, however small, it is possible to find d such thatf(c) - e < f(c) < f(c) + efor ALL x in c - d < x < c + d.In simpler terms, it is possible to find an interval around x such that for ALL values of x' in that interval, the value of the function, f(x'), is close to f(x).Determining continuity visually, it is easy: if the function can be drawn without lifting your pencil, then it is continuous and if you cannot, it is not.
Yes.
A variable defined on a continuous interval as opposed to one that can take only discrete values.
If the function is continuous in the interval [a,b] where f(a)*f(b) < 0 (f(x) changes sign ) , then there must be a point c in the interval a<c<b such that f(c) = 0 . In other words , continuous function f in the interval [a,b] receives all all values between f(a) and f(b)
Yes, it is a Continuous variable measured along an equidistant scale.
why doesn't wiki allow punctuation??? Now prove that if the definite integral of f(x) dx is continuous on the interval [a,b] then it is integrable over [a,b]. Another answer: I suspect that the question should be: Prove that if f(x) is continuous on the interval [a,b] then the definite integral of f(x) dx over the interval [a,b] exists. The proof can be found in reasonable calculus texts. On the way you need to know that a function f(x) that is continuous on a closed interval [a,b] is uniformlycontinuous on that interval. Then you take partitions P of the interval [a,b] and look at the upper sum U[P] and lower sum L[P] of f with respect to the partition. Because the function is uniformly continuous on [a,b], you can find partitions P such that U[P] and L[P] are arbitrarily close together, and that in turn tells you that the (Riemann) integral of f over [a,b] exists. This is a somewhat advanced topic.
Acceleration = (speed at the end of some time interval minus speed at the beginning of the interval)/(length of the time interval)
The contour interval
Notice the pattern around that contour line. Then determine the interval that the surrounding contour lines are increasing or decreasing by. Ex. 50 100 150 200, the contour interval would then be 50
sorry but are gone mad
Interval training is periods of work followed by periods of rest. This is known as work:rest ratio. This is commonly used to train the anaerobic energy system. Continuous training, of which there are many forms does not involve rest periods, although it could involve periods of different intensities (such as Fartlek training).
On the request of the owner a continuous survey may be carried out on the hull which all compartment of the hull are opened for survey and testing in rotation, with a five year interval between examination of each part.
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