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Please explain to me whether the statement below are true or false:
1. If F(x) is an antiderivative of f(x), then y = F(x) is a solution to the
differential equation dy/dx = f(x).
2. If y =F(x) is a solution to the differential equation dy/dx = f(x),
then F(x) is an antiderivative of f(x)
3. If an object has constant nonzero acceleration, then the position
of the object as a function of time is a quadratic polynomial.

Thanks for all the help!

2006-11-18 06:33:39 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

All are true. For the last one,
suppose acceleration = a.
Then the velocity is at + c
and the position is at^2/2 + ct + d,
where c and d are constants.

2006-11-18 06:44:48 · answer #1 · answered by steiner1745 7 · 0 0

that's actual and that i'm imparting you with the completed information for it from between the web content on the internet that can assist you you in comprehend-how. In arithmetic, a sq. variety, each and every so often called a perfect sq., is an integer that is written because of the fact the sq. of another integer; in different words, it relatively is the made of a few integer with itself. So, as an occasion, 9 is a sq. variety, because of the fact it relatively is written as 3 × 3. sq. numbers are non-adverse. yet in any different case of asserting that a (non-adverse) variety is a sq. variety, is that its sq. root is back an integer. as an occasion, ?9 = 3, so 9 is a sq. variety. a rather good integer that has no perfect sq. divisors different than a million is named sq.-unfastened. the standard notation for the formulation for the sq. of a variety n isn't the product n × n, however the equivalent exponentiation n2, often stated as "n squared". For a non-adverse integer n, the nth sq. variety is n2, with 02 = 0 being the zeroth sq.. the considered sq. would be prolonged to a pair different variety structures. If rational numbers are blanketed, then a sq. is the ratio of two sq. integers, and, conversely, the ratio of two sq. integers is a sq. (e.g., 4/9 = (2/3)2). commencing with a million, there are sq. numbers as much as and which contains m.

2016-10-15 17:22:39 · answer #2 · answered by ? 4 · 0 0

1.true

2.true

3.true

2006-11-18 06:39:44 · answer #3 · answered by raj 7 · 0 0

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