Proofs

Because the gypsum only needs 2 percent added to the mix

It is because the partial fractions are simply another way of expressing the same algebraic fraction.

The worst case occurs when data is already sorted where the complexity is O(n^2) instead of the well known O(n log n)

None.

There is nothing in the question that requires the polygon to be regular and it is difficult to define a radius, in any meaningful way, for an irregular shape.

A statement in maths is true only if it is proved by a series of mathematical manipulations and logic for any GENERAL number for which the statement should satisfy. Otherwise, it is only a conjecture. This is the beauty of mathematics and it is proof which differentiate mathematics from all other fields.

This is used due to the extreme simplicity of binary numbers (take a look at the addition table, and at the multiplication table, for binary numbers). This makes it much simpler to design circuits to do the calculations. Also, it is usually simpler, and safer, to distinguish two different states (for example, a high and a low voltage, to represent 1 and 0, respectively), than to distinguish ten different states.

There are infinity square roots. Here is why. 1s square root is one, 4s is 2, and goes on forever. So, any whole number is a square root.

My guess. I am only in middle school.

He didn't write it. What he did was to write in the margin of a book that he had a proof but there was not enough space to write it there.

no

Argument

a^2 + b^2 + c^2 - ab - bc - ca = 0=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0

=> a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2 = 0

=> (a - b)^2 + (b - c)^2 + (c - a)^2 = 0

Each term on the left hand side is a square and so it is non-negative.

Since their sum is zero, each term must be zero.

Therefore:

a - b = 0 => a = b

b - c = 0 => b = c.

You might mean a halocline - that's when there's a great enough difference in salinity that the water separates into two phases, with a surface between them.

About 97.

The Answers community requires more information for this question. Please edit your question to include more context. Core of WHAT?

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.

More Than Friends, Less Than Lovers - One Way featuring Al Hudson

non square rhombus

yes. they are not congruent

Yes.

I'm assuming this is talking asking about boolean logic (the question makes little sense otherwise).

If a and b are equal, then the complement of a and the complement of b are equal.

yes they did the goverment did it for there own personal gain

If we assume that the sqare root of 5 is a rational number, then we can write it as a/b in its simplest form, where a and b have no common factors.

Therefore 5 = a2/b2

Therefore 5b2 = a2

Therefore a2 is divisible by 5, because b2 is an integer

Therefore a is divisible by 5, because 5 is a prime number.

Therefore 5 = 5c/b, where c is an integer

Therefore 1 = c/b

Therefore c = b

Therefore sqrt 5 = 5c/c = 5, which is impossible.

So sqrt5 cannot be expressed in the form a/b, and is irrational.

An Experiment has no use without its documentation. Regarding this: The Lab Book; it contains - Protocol; Equipment; chemicals this and procedures that, compression of the results occurs in the Conclusion, while The Discussion is the Section of the Report that follows the Conclusion.

{An experiment has no use without its documentation. Re: the Lab Book : Protocol; Equipment; chemicals this and procedures that, discussion of the results occurs in the Conclusion}.

The proof is by contradiction: assume there is a finite number of prime numbers and get a contradiction by requiring a prime that is not one of the finite number of primes.

Suppose there are only a finite number of prime numbers.

Then there are n of them.; and

they can all be listed as: p1, p2, ..., pn in order with there being no possible primes between p(r) and p(r+1) for all 0 < r < n.

Consider the number m = p1 × p2 × ... × pn + 1

It is not divisible by any prime p1, p2, ..., pn as there is a remainder of 1.

Thus either m is a prime number itself or there is some other prime p (greater than pn) which divides into m.

Thus there is a prime which is not in the list p1, p2, ..., pn.

But the list p1, p2, ..., pn is supposed to contain all the prime numbers.

Thus the assumption that there is a finite number of primes is false;

ie there are an infinite number of primes.

QED.