answer:Given that the ratio of the areas of three triangles (area scale factor) is {eq}4:9:16 {/eq}, then the linear scale factor (ratio of perimeters) is equal to begin{align} sqrt{4}:sqrt{9}:sqrt{16}=2:3:4 end{align} This means that if the perimeter of the largest triangle is {eq}P {/eq}, then the perimeter of the second triangle is begin{align} P_2=frac{3}{4}P end{align} And the perimeter of the smallest triangle is begin{align} P_3=frac{1}{2}P end{align} Therefore, if the perimeter of the largest triangle is 60 cm, then the perimeter of the second triangle is begin{align} P_2&=frac{3}{4}times 60; rm cm[0.3cm] &= boxed{color{blue}{45; rm cm}} end{align} And the perimeter of the smallest triangle is begin{align} P_3&=frac{1}{2}times 60; rm cm[0.3cm] &= boxed{color{blue}{30; rm cm}} end{align}

answer:To determine which element is oxidized and which is reduced, we need to assign oxidation states to all atoms in the reaction. In its standard state, chlorine has an oxidation state of 0. In the reactants, chlorine has an oxidation state of 0 in Cl2. In the products, chlorine has an oxidation state of -1 in NaCl. Therefore, chlorine is reduced because its oxidation state decreases. Bromine has an oxidation state of -1 in its standard state. In the reactants, bromine has an oxidation state of -1 in NaBr. In the products, bromine has an oxidation state of 0 in Br2. Therefore, bromine is oxidized because its oxidation state increases. In the given redox reaction, chlorine (Cl) is reduced and bromine (Br) is oxidized.

answer:No, Parseval's identity applies specifically to Hilbert spaces, such as L^2, but not in a general Banach space, like L^1. While Parseval's identity shows the Riemann-Lebesgue lemma for L^2, it cannot be directly extended to L^1 without additional arguments. An alternative proof, often referred to as one that "avoids step functions," involves a u-substitution. This leads to: hat{f}(xi) = int_{mathbb{R}} f(x)e^{ixxi} dx = int_{mathbb{R}} f(x + frac{pi}{xi})e^{ixxi}e^{i pi} dx = - int_{mathbb{R}} f(x + frac{pi}{xi})e^{ixxi} dx By averaging these two forms, we obtain: hat{f}(xi) = frac{1}{2} int_{mathbb{R}} (f(x) - f(x + pi/xi)) e^{ixxi} dx Since |e^{ix}| = 1, we can then deduce: hat{f}(xi) leq frac{1}{2} int_{mathbb{R}} |f(x) - f(x+pi/xi)| dx As xi to 0, the proof is completed, assuming the limit and integral can be interchanged. However, demonstrating this interchange typically requires arguments involving step functions, thus the term "avoid step functions" is used in a relative sense.

answer:To show that alpha is not monic if and only if there is a non-zero beta neq sigma_0 with alpha beta = sigma_0, we proceed with two implications: 1. (implies): Assume that alpha is not monic. By definition, a linear transformation is monic if its kernel is trivial, i.e., ker(alpha) = {0}. Since alpha is not monic, there exists a non-zero vector v_0 in V with alpha(v_0) = 0. Define beta to be the endomorphism that maps every basis vector to v_0. Consequently, for any v in V, we have alpha beta(v) = alpha(cv_0) = calpha(v_0) = 0, where c in F is a scalar. Thus, alpha beta = sigma_0 and beta neq sigma_0 because beta(v_0) = v_0 neq 0. 2. (impliedby): Conversely, suppose alpha beta = sigma_0 for some beta neq sigma_0. Assume, to reach a contradiction, that alpha is monic. Then, ker(alpha) = {0}, and for all v in V, alpha(beta(v)) = 0 implies beta(v) = 0. However, this would mean beta = sigma_0, which contradicts the assumption that beta neq sigma_0. Hence, alpha must not be monic. In summary, alpha is not monic if and only if there exists an endomorphism beta neq sigma_0 such that alpha beta = sigma_0.