Here are the newest items added to Quandaries & Queries:
start over

		Page 660/886

	BEDMAS		2005-09-01
		From A student: I am a student and am wondering about the answer to this question. 56/2(31) is the answer 7? Answered by Harley Weston.
	A typical farm		2005-09-01
		From Pagedi: A typical farm is about 700 acres. How many square miles is this? Please show me the work so, that I will know I to perform this problem myself. Answered by Penny Nom.
	Metric conversion		2005-09-01
		From Donna: I know there is an equation to figuring out, for example. .5526 km = _______ mm Answered by Penny Nom.
	The sum of a series		2005-08-31
		From Aamod: Find the sum of the given series till n terms: (14\1*3) + (24\3*5) + (34\5*7)............... till n terms Answered by Chri Fisher.
	12-42, 8-26, 10-34, 9-30, 16-58		2005-08-31
		From Elizabeth: here is a math question my niece gave me, grade 7 12-42, 8-26, 10-34, 9-30, 16-58 first day of school and this is what they give her no explanation Answered by Harley Weston.
	1,4,9,1,6,2,5,3,6,4,9,6,4,8,1		2005-08-30
		From Liz: Find the next four numbers to the sequence 1,4,9,1,6,2,5,3,6,4,9,6,4,8,1,___,___,___,___. Answered by Penny Nom.
	cos x * cos 2x * cos 4x * cos 8x		2005-08-29
		From Leandro: A = cos x * cos 2x * cos 4x * cos 8x What's the value of log A at base 2? Answered by Chris Fisher and Penny Nom.
	The area of a lot		2005-08-29
		From Richard: My wife and I are interested in buying property in Idaho but the owner can't give us a square footage of the lot. The dimensions are as follows: 121.0 on the left side 157.0 on the right side 135.0 on the bottom 162.0 on the top The bottom right corner of the lot is a true right angle, the rest are not. Answered by Penny Nom.
	The length of a lot		2005-08-28
		From Billy: If I have 3 acres and the front is 300 ft across how many feet would I go down the sides to equal 3 acres? Answered by Penny Nom.
	1_2_3_4_5_6_7_8_9 = 1: Fill in the blanks		2005-08-26
		From James: Try to replace the blanks below with + or - to make the statement correct 1_2_3_4_5_6_7_8_9 = 1 Answered by Penny Nom.
	Graphing a linear inequality		2005-08-26
		From Gina: When graphing a linear inequality, how do you know if the inequality represents the area above the line? Answered by Penny Nom.
	y = log(x) + x. Solve for x.		2005-08-26
		From Alain: I have the following equation: y = ln(x) + x How do I solve for x? Answered by Penny Nom.
	The area of a triangle		2005-08-26
		From Martha: I have a triangle that with a base of 60' and the two sides of 37'. I know the formula for area is A=1/2 (b*h) but how do I find the height?? Answered by Penny Nom.
	The area of a pentagon		2005-08-24
		From Susan: I need to calculate the square footage of a regular pentagon (all angles are equal). Each side is 16 feet long. Answered by Penny Nom.
	Labour efficiency		2005-08-23
		From Rob: The problem, on the surface, seems very simple and yet has created some controversy among a group of accountants. The problem itself has to do with labour efficiency rates and only involves two variables; standard working hours, and actual working hours. The difficulty lies in deriving an efficiency % from these two numbers. Standard working hours or the targeted number of labour hours required to produce one widget, which I will represent as "s". Actual working hour or the actual number of labour hours require to produce one widget, which I will represent as "a". Labour efficiency I will represent with "E". The prevailing calculation with which I have a problem with is this: s/a=E or if s=3000, and a=4000 then 3000/4000=75% What bothers me about the calculation is that the standard hours get represented as a percentage of the actual hours and in my opinion changes the focus of the calculation from standard or target, where it should be, to the actual hours. I cannot define why, but this just seems inherently wrong to me. The calculation that I use: (1+((s-a)/s))=E or if s=3000, and a=4000 then (1+((3000-4000)/3000))=66.67% My calculation is like a %change from standard calculation. However, there is something that also concerns me about my calculation. If you substitute 100 for a and 50 for s, then you come to a quandary, because if you plug those numbers into the second equation the result is of course zero % efficient which doesn't sit right with me either. If you plug them into the first calculation you get 50% efficiency which doesn't really seem to work either, because you require 100% more hours to do the same work in this case. ??? Is the first calculation correct? Am I missing something altogether? Are both calculations off base? Answered by Harley Weston.

		Page 660/886