answer:The symmetry in the resistive network of an op amp configuration can be intuitively understood by visualizing the circuit as a voltage divider or a seesaw. The voltage at point ( text{x} ) is determined by the relative magnitudes of the resistors and the input voltages. In Figure 1, the voltage at ( text{x} ) can be determined graphically or mentally as the voltage that creates a balance between the voltage drops across ( R_1 ) and ( R_2 ), considering their relative sizes. For the values given in the original question, ( V(x) ) would be 1/3 of the way between 1 V and 10 V, resulting in 4 V. Analogously, Figure 2 illustrates the voltage seesaw analogy for an inverting op-amp configuration. Here, the voltage at the input node (virtual ground) is determined by the ratio of the resistor values. As the input voltage ( V_1 ) changes, the output voltage adjusts to maintain the virtual earth condition, effectively acting as a lever with arms proportional to ( R_1 ) and ( R_2 ). In summary, the symmetry arises because the op amp strives to maintain a zero voltage difference across its inputs, causing the voltage at ( text{x} ) to be a linear combination of the input voltages, weighted by the respective resistor values. This behavior allows for precise voltage control and is the basis for various op amp circuit configurations. [](https://i.stack.imgur.com/xv8Zv.png) [](https://i.stack.imgur.com/4zqOu.png)

answer:Using the Lagrangian from the original answer, we can derive the first-order conditions with the new utility function: {eq}c: l - a = 0 l: c - b = 0 n: aw(1-tau) - b = 0 {/eq} Solving for a and b, we get: {eq}a = l b = c {/eq} Substituting these into the third first-order condition, we get: {eq}lw(1-tau) = c {/eq} Since c = b = l, we can solve for the optimal labor supply: {eq}n^* = frac{1}{w(1-tau)} {/eq} Therefore, the consumer's optimal labor supply is inversely related to the tax rate and wage, and is not affected by profit.

answer:To evaluate the integral {eq}displaystyle int_0^4 |x - 3| , dx{/eq} with the Fundamental Theorem of Calculus, consider the piecewise definition of the absolute value: {eq}displaystyle |x-3|=begin{cases} x-3, & x geq 3 -x+3, & x < 3 end{cases}{/eq} Splitting the integral at {eq}x = 3{/eq}, we have: {eq}displaystyle int_0^4 |x - 3| , dx = int_0^3 (-x + 3) , dx + int_3^4 (x - 3) , dx{/eq} Now, compute each integral separately: {eq}begin{align*} int_0^3 (-x + 3) , dx &= left[-frac{1}{2}x^2 + 3xright]_0^3 &= left[-frac{1}{2}(3)^2 + 3(3)right] - left[-frac{1}{2}(0)^2 + 3(0)right] &= left[-frac{9}{2} + 9right] - 0 &= frac{9}{2} end{align*}{/eq} {eq}begin{align*} int_3^4 (x - 3) , dx &= left[frac{1}{2}x^2 - 3xright]_3^4 &= left[frac{1}{2}(4)^2 - 3(4)right] - left[frac{1}{2}(3)^2 - 3(3)right] &= left[frac{16}{2} - 12right] - left[frac{9}{2} - 9right] &= 4 - left(-frac{9}{2}right) &= 4 + frac{9}{2} &= frac{17}{2} end{align*}{/eq} Adding the results from both integrals: {eq}displaystyle frac{9}{2} + frac{17}{2} = frac{26}{2} = 13{/eq} Therefore, the value of the integral is {eq}boxed{13}.{/eq}

answer:While saliva and urine both have some common components, like urea and electrolytes, they are distinct bodily fluids with different functions. Saliva is primarily composed of water, electrolytes, proteins, enzymes, and antimicrobial factors. A small amount of urea, a byproduct of protein metabolism, can be found in saliva, but this does not mean that saliva contains urine. Urine, on the other hand, is a much more complex mixture of waste products, including urea, creatinine, uric acid, enzymes, hormones, and various inorganic ions. The presence of nitrate in saliva, which can be derived from unused dietary nitrate, does not classify saliva as containing urine. Therefore, while there are some similarities, human saliva is not a form of urine and should not be confused with it.