answer:The Sumerians, an ancient civilization that flourished in Mesopotamia (modern-day Iraq) from around 3500 to 1750 BC, developed cuneiform, one of the earliest known forms of writing. Cuneiform, which means "wedge-shaped" in Latin, was a system of writing that used wedge-shaped marks made on clay tablets. The Sumerians used a stylus, a sharp-pointed tool, to press into the soft clay, creating wedge-shaped marks that represented words, syllables, or even entire concepts. These clay tablets were then baked in a kiln to harden them and preserve the writing. Clay tablets

answer:100%, as the Stamp Act was the only act repealed completely, and it accounted for all of the taxes that were repealed.

answer:Question 1: Yes, the explanation for single-photon interference in a double-slit experiment is generally correct. The photon's wavefunction splits as it passes through the two slits, and these wavefunctions interfere with each other, creating an interference pattern on the screen. Question 2: For two photons passing through the double slit simultaneously, each photon's wavefunction still splits and interferes with itself, creating its own interference pattern. However, the interference between the photons themselves is negligible, meaning that their probability waves do not directly interfere with each other. Each photon behaves independently, and the overall interference pattern on the screen is a superposition of the patterns created by each photon.

answer:To prove the identity, consider the binomial theorem, which states that for any non-negative integer n and any real numbers a and b: (a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n-k} b^k Let a = b = 1 to get: 2^n = sum_{k=0}^{n} binom{n}{k} Now, we use the property that k binom{n}{k} = n binom{n-1}{k-1}, valid for n > 0 and k > 0. This can be derived from the definition of binomial coefficients: binom{n}{k} = frac{n!}{k!(n-k)!} Differentiating both sides of the binomial theorem with respect to x when a = x + 1 and b = 1 gives: n(x+1)^{n-1} = sum_{k=0}^{n} k binom{n}{k} x^{k-1} Plugging x = 1: n(1+1)^{n-1} = sum_{k=0}^{n} k binom{n}{k} = n cdot 2^{n-1} Alternatively, consider a combinatorial interpretation: suppose we have a set of n elements and want to choose a subset with one designated element. We can first choose the designated element (in n ways) and then choose a subset from the remaining n-1 elements (in 2^{n-1} ways). Thus, the total number of ways is n cdot 2^{n-1}. We can also prove the identity algebraically. Let's manipulate the sum: sum_{k=0}^{n} k binom{n}{k} = sum_{k=0}^{n} frac{n!}{k!(n-k)!} cdot k = sum_{k=0}^{n} frac{n cdot n!}{k!(n-k)!} = n sum_{k=0}^{n} binom{n-1}{k-1} Make the substitution m = k - 1: n sum_{m=0}^{n-1} binom{n-1}{m} = n cdot 2^{n-1} Hence, we have shown the identity in multiple ways: through differentiation, combinatorial interpretation, and algebraic manipulation.