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Here are the newest items added to Quandaries & Queries: |
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start over |
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BEDMAS |
2005-09-01 |
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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. |
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A typical farm |
2005-09-01 |
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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. |
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Metric conversion |
2005-09-01 |
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From Donna: I know there is an equation to figuring out, for example. .5526 km = _______ mm Answered by Penny Nom. |
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The sum of a series |
2005-08-31 |
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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. |
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12-42, 8-26, 10-34, 9-30, 16-58 |
2005-08-31 |
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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. |
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1,4,9,1,6,2,5,3,6,4,9,6,4,8,1 |
2005-08-30 |
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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. |
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cos x * cos 2x * cos 4x * cos 8x |
2005-08-29 |
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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. |
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The area of a lot |
2005-08-29 |
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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. |
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The length of a lot |
2005-08-28 |
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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. |
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1_2_3_4_5_6_7_8_9 = 1: Fill in the blanks |
2005-08-26 |
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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. |
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Graphing a linear inequality |
2005-08-26 |
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From Gina: When graphing a linear inequality, how do you know if the inequality represents the area above the line? Answered by Penny Nom. |
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y = log(x) + x. Solve for x. |
2005-08-26 |
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From Alain:
I have the following equation:
y = ln(x) + x
How do I solve for x?
Answered by Penny Nom. |
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The area of a triangle |
2005-08-26 |
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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. |
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The area of a pentagon |
2005-08-24 |
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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. |
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Labour efficiency |
2005-08-23 |
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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. |
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