answer:To demonstrate uniform convergence, we need to show that for any epsilon > 0, there exists an N in mathbb{N} such that for all n > N and for all x in [0, 1], we have |f_n(x) - f(x)| < epsilon. Given f_n(x) = frac{1}{x+n} and f(x) = 0, we can write: |f_n(x) - f(x)| = left|frac{1}{x+n}right| Since the sequence is continuous for all n and x in [0, 1], the limit as n approaches infinity is f(x) = 0. However, to prove uniform convergence, we need to find an N that works independently of x. The sequence's supremum on [0, 1] occurs at x = 0 because the function is decreasing as x increases. Thus: s_n = sup_{x in [0, 1]} |f_n(x) - f(x)| = sup_{x in [0, 1]} left|frac{1}{x+n}right| = left|frac{1}{n}right| As n approaches infinity, s_n to 0. Therefore, for any epsilon > 0, we can choose N such that n > N implies frac{1}{n} < epsilon, which is equivalent to n > frac{1}{epsilon}. This N works for all x in [0, 1], establishing uniform convergence.

answer:The LED current in the MP3202-based circuit is primarily controlled by resistor R7. To reduce the maximum current to 60mA, you should calculate the new resistance value for R7. The formula to calculate the resistor value for a constant current is: [ R = frac{V_{FB}}{I_{LED}} ] Where ( V_{FB} ) is the feedback voltage (104mV in this case) and ( I_{LED} ) is the desired LED current (60mA). [ R = frac{0.104V}{0.06A} ] [ R approx 1.733Omega ] Considering the tolerance of the chip's Vref and the resistor, it's recommended to choose a slightly higher value for R7, such as 1.8 ohms, to ensure the current stays below 60mA. The input signal from the microcontroller will continue to control the LED dimming.

answer:The exponential function 5^x is always positive for any real number x. Therefore, it can never be equal to zero.

answer:The Lewis structure of water ({eq}rm H_2O {/eq}) represents the bonding and lone electron pairs in the molecule. Oxygen (O), with 6 valence electrons, forms double covalent bonds with each of the two hydrogen (H) atoms, which have 1 valence electron. To complete its octet, oxygen shares two pairs of electrons, while each hydrogen shares one pair, resulting in a total of two single covalent bonds. The remaining two electron pairs on the oxygen atom are non-bonding (lone pairs), giving the molecule a bent shape. Here is the Lewis structure of water: <img src="https://latex.codecogs.com/png.image?&space;inline&space;rm&space;H&space;-overset{large&space;{color{red}{2}}}{overset{large&space;{color{blue}{cdot}}&space;cdot}{O}}&space;-&space;H" title="rm H -overset{large {color{red}{2}}}{overset{large {color{blue}{cdot}} cdot}{O}} - H" />