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Question:Example: Cell A1 = 4:00 AM. I'd like Cell A2 to show 8:00 PM (that's cell "A18" hours). I'd like Cell A3 to show 12:00 AM (that's cell A14 hours). How do I set up the formula's in cells A2 and A3?
Note: the formula
=TIME(HOUR(A1)+16, MINUTE(A1), SECOND(A1))
results in a #VALUE! error excelerate and Ted Pack  my thanks to both. excelerate you were the first correct answer; but I goofed in applying your first answer.
Answers:TIME(HOUR(A1)+16, MINUTE(A1), SECOND(A1)) worked fine for me when I had A1 as ="04:00" (Note the equal sign and quotation marks.)
Answers:TIME(HOUR(A1)+16, MINUTE(A1), SECOND(A1)) worked fine for me when I had A1 as ="04:00" (Note the equal sign and quotation marks.)
Question:Using a balance scale and weights of 1kg, 3kg, 9kg, and 27kg, how many objects of different weights can be weighed? The objects to be weighed and the wights may be placed in either pan of the balance scale.
(show work) Make sure to answer all parts of question. When someone gets the answer right and shows work i will post the correct answer and select best answer. HAVE FUN
Answers:1 2(31) 3 4(3+1) 5(931) 6(93) 7(9+13) 8(91) 9 10(9+1) 11(9+31) 12(9+3) 13(9+3+1) 14(27931) 15(2793) 16(27+193) 17(2791) 18(279) 19(27+19) 20(27+319) 21(27+39) 22(27+3+19) 23(2731) 24(273) 25(27+13) 26(271) 27 28>40 add 27 to the instructions for 1>13 a minus sign means put the weight in the same pan as the object being measured. You can weigh 40 different weights, or 41 if you include 0kg
Answers:1 2(31) 3 4(3+1) 5(931) 6(93) 7(9+13) 8(91) 9 10(9+1) 11(9+31) 12(9+3) 13(9+3+1) 14(27931) 15(2793) 16(27+193) 17(2791) 18(279) 19(27+19) 20(27+319) 21(27+39) 22(27+3+19) 23(2731) 24(273) 25(27+13) 26(271) 27 28>40 add 27 to the instructions for 1>13 a minus sign means put the weight in the same pan as the object being measured. You can weigh 40 different weights, or 41 if you include 0kg
Question:1. 54+2 24 6
2. 5 272 90+ 40 48
Answers:1) 54 + 2 24  6 = (3^2 * 6) + 2 (2^2 * 6)  6 = 3 6 + 2(2) 6  6 = 3 6 + 4 6  6 = 7 6  6 = 6 6 2) 5 27  2 90 + 40  48 = 5 (3^2 * 3)  2 (3^2 * 10) + (2^2 * 10)  (4^2 * 3) = 5(3) 3  2(3) 10 + 2 10  4 3 = 15 3  6 10 + 2 10  4 3 = 15 3  4 3  4 10 = 11 3  4 10
Answers:1) 54 + 2 24  6 = (3^2 * 6) + 2 (2^2 * 6)  6 = 3 6 + 2(2) 6  6 = 3 6 + 4 6  6 = 7 6  6 = 6 6 2) 5 27  2 90 + 40  48 = 5 (3^2 * 3)  2 (3^2 * 10) + (2^2 * 10)  (4^2 * 3) = 5(3) 3  2(3) 10 + 2 10  4 3 = 15 3  6 10 + 2 10  4 3 = 15 3  4 3  4 10 = 11 3  4 10
Question:whats the answer if you subtract the two functions, can you also show how it is worked out
thanks
x+ 2y
______
x^2 + 4xy +4y^2
minus
2x
___
x^2 4y^2
Answers:1. factorise everything (x+ 2y) / (x^2 + 4xy +4y^2)  (2x) / (x^2 4y^2) = (x+ 2y) / (x + 2y)(x +2y)  (2x) / (x  2y)(x + 2y) in this case simplify the first term = 1/(x +2y)  (2x) / (x  2y)(x + 2y) multiply the first term by (x  2y) to make a common denominator =(x  2y)/(x  2y)(x +2y)  (2x) / (x  2y)(x + 2y) now subtract the numerators =[(x  2y)  (2x)] / (x  2y)(x + 2y) =[ x  2y] / (x  2y)(x + 2y) =  ( x+ 2y) / (x  2y)(x + 2y) divide the common factor = 1 / (x  2y) hope that helps Good Luck!
Answers:1. factorise everything (x+ 2y) / (x^2 + 4xy +4y^2)  (2x) / (x^2 4y^2) = (x+ 2y) / (x + 2y)(x +2y)  (2x) / (x  2y)(x + 2y) in this case simplify the first term = 1/(x +2y)  (2x) / (x  2y)(x + 2y) multiply the first term by (x  2y) to make a common denominator =(x  2y)/(x  2y)(x +2y)  (2x) / (x  2y)(x + 2y) now subtract the numerators =[(x  2y)  (2x)] / (x  2y)(x + 2y) =[ x  2y] / (x  2y)(x + 2y) =  ( x+ 2y) / (x  2y)(x + 2y) divide the common factor = 1 / (x  2y) hope that helps Good Luck!
From Youtube
Subtracting integers video  three models to explain how subtraction of negative numbers works :I explain three different models you can use to show why subtraction of integers works: 1) number line jumps; 2) concept of difference; 3) positive/negative counters. The bottom line is that students need to learn the shortcuts and rules, but these models can help justify the rules.
adding and subtracting decimals :in this video i show you how to add and subtract decimals and the methods are very similar to whole numbers and you can add 3 or more decimals togther just like with whole numbers i give plenty of examples and a challengeing problem at the end where i use logic to work out the problem you probably not get a problem like that on an exam always good to check these with a calculator while learning