tarimshahab

I'm not sure if you want binary fractions, or a binary number with a 'decimal' point. If you want a binary fraction, then just change the numerator and denominator into binary. Since 1(dec)=1(bin) and 2(dec)=10(bin): 1/2(dec) = 1/10(bin) Similarily 1/4(dec)=1/100(bin). If you want to find the equivalent binary number with a 'decimal' point, then there are two ways to do it: First way using the number 3/4=0.75 as an example. 1 ) Multiply the dec# by a power of two so that it is integral: ex. 0.75*2^2=3 2 ) Convert this integral number into binary: ex. 3(dec)=11(bin) 3 ) Now divide this new binary number by the same power of 2, you multiplied the decimal one by. Since you are in binary, dividing by a power of two is just moving the 'decimal' point: ex 11(bin) / 2^2(dec) = 0.11 (moving the decimal twice to the left). You have your answer. 3/4=0.75(dec)= 0.11(bin) Second way using the same example 1 ) Multiply decimal fraction by 2: ex. 0.75*2= 1.5 2 ) Take the number in the unit place (one's place): ex. in this case it would be a 1, this is the first digit after the decimal point for the binary equivalent. ie. 0.75(dec)=0.1????(bin) 3 ) Take the fraction part of the new decimal number and multiply by two again: ex. 0.5*2= 1.0 4 ) Again take the number in the units place as the next binary number: so 0.75(dec)=0.11???(bin) 5 ) Keep repeating this until you get a 0 as the fraction part of the decimal number, or until the decimal part starts repeating: in this case we got a 0 in the fractions part of the decimal number 1.0, so we stop. If the fraction part starts repeating you know that the binary fraction repeats too.

All your answers are wrong, but not because of your derivation, but for the same mistake you make; probably because you were trying to do it quickly. Specifically, generally speaking, (x+a)^b IS NOT EQUAL TO x^b+a^b So basically for the lesson 5 question: 6/(x+4)^2 doesn't simplify to 6/(x^2+16) Same for the rest of your questions. Fix this, and I believe you'll get the right answer, since you don't make any other mistake. BTW, if you want to quickly check if you got the right answer then the way to go is: http://www.wolframalpha.com/ Just input "derive f(x)", you can derive and integrate and do a bunch of other stuff.

I'm not sure exactly what you are asking. For your pendulum experiment, yes you could use that to find the acceleration due to gravity by using the following pendulum equation: T = 2*pi*sqrt(L/g). where T is the period of the swing, L is the length of the pendulum, and g is the acceleration of gravity Isolated for g gives: g = L*(2*pi/T)^2 Once you know g, you can use weight = m*g, to find the weight on different plantets. If however, you know that the gravity is 1/6 the gravity on Earth, then you can just measure the weight on Earth, and divide by 6.

For Naming The new system you mentioned, is ONLY used for ionic substances, and replaces as you say the old ferric/ferrous system. Note also, that eihter of these systems is only needed for atoms that can exist with more than on oxidation number so basically the transition metals. I'll explain the new system by an example: Fe2O3 . So start by naming the metal, in this case iron. Then you have to figure out the oxidation number of the iron in the compound. You know that oxygen is almost always has oxidation number of -2. (Except in peroxides, in which it is -1). Anyway, so O is -2, and there are 3 of them, which gives a imaginary charge of -2*3= -6 coming from the oxygen. Since the whole compound is neutral, the -6 must be neutralized by the iron. So -6 / 2 = +3. There fore the oxidation number on the iron is +3. Once you have the oxidation number on the metal you put it as roman numerals in brackets, without a space: iron(III) Now just put the non-metal part in: iron(III) oxide. Similarily CuSO4 is copper(II) sulfate, because the oxidation of copper is +2. For hydrates such as, CuSO4 · 5H2O, you must also name the water part. Just do this by using the greek prefixes: mono, di, tri, tetra, penta, etc. So CuSO4 · 5H2O is copper(II) sulfate pentahydrate. Solubility in Water There are rules for these you will just have to memorize. The rules do have some logic to them, but memorizing is just easier. Here is a site with the rules. http://www.chem.sc.edu/faculty/morgan/resources/solubility/