Understanding the Concept of Sum in Physics: A Beginner's Guide

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The discussion focuses on the propagation of uncertainties in physics, specifically using the formula Δz = √((Δa)² + (Δb)²) for the sum of two measurements. It emphasizes the need to convert percentage uncertainties into absolute uncertainties before applying the formula. The original poster was banned for not demonstrating effort in their homework submissions despite receiving multiple warnings. The thread has been locked to prevent further discussion. Understanding these principles is crucial for accurate calculations in physics.
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Homework Statement
Find the desired sum: (3±4%)+(4±0.01)
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please help me
 
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Well, why not just use the formula for propagation of uncertainties, if ##z = a + b##, then$$\Delta z = \sqrt{(\Delta a)^2 + (\Delta b)^2}$$N.B. you need to convert the percentage uncertainty in the first measurement into an absolute uncertainty first.
 
OP has been banned for repeatedly posting homework with zero effort shown, even after multiple warnings. Have a nice day. Thread is locked.
 
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Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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