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BEDMAS 2005-09-01
A student pose la question :
I am a student and am wondering about the answer to this question.

56/2(31)

is the answer 7?


Harley Weston lui répond.
A typical farm 2005-09-01
Pagedi pose la question :
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.
Penny Nom lui répond.
Metric conversion 2005-09-01
Donna pose la question :
I know there is an equation to figuring out, for example. .5526 km = _______ mm
Penny Nom lui répond.
The sum of a series 2005-08-31
Aamod pose la question :
Find the sum of the given series till n terms:

(14\1*3) + (24\3*5) + (34\5*7)............... till n terms

Chri Fisher lui répond.
12-42, 8-26, 10-34, 9-30, 16-58 2005-08-31
Elizabeth pose la question :
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

Harley Weston lui répond.
1,4,9,1,6,2,5,3,6,4,9,6,4,8,1 2005-08-30
Liz pose la question :
Find the next four numbers to the sequence 1,4,9,1,6,2,5,3,6,4,9,6,4,8,1,___,___,___,___.
Penny Nom lui répond.
cos x * cos 2x * cos 4x * cos 8x 2005-08-29
Leandro pose la question :

A = cos x * cos 2x * cos 4x * cos 8x

What's the value of log A at base 2?


Chris Fisher and Penny Nom lui répond.
The area of a lot 2005-08-29
Richard pose la question :
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.

Penny Nom lui répond.
The length of a lot 2005-08-28
Billy pose la question :
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?
Penny Nom lui répond.
1_2_3_4_5_6_7_8_9 = 1: Fill in the blanks 2005-08-26
James pose la question :
Try to replace the blanks below with + or - to make the statement correct
1_2_3_4_5_6_7_8_9 = 1

Penny Nom lui répond.
Graphing a linear inequality 2005-08-26
Gina pose la question :
When graphing a linear inequality, how do you know if the inequality represents the area above the line?
Penny Nom lui répond.
y = log(x) + x. Solve for x. 2005-08-26
Alain pose la question :

I have the following equation:

y = ln(x) + x

How do I solve for x?


Penny Nom lui répond.
The area of a triangle 2005-08-26
Martha pose la question :
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??
Penny Nom lui répond.
The area of a pentagon 2005-08-24
Susan pose la question :
I need to calculate the square footage of a regular pentagon (all angles are equal). Each side is 16 feet long.
Penny Nom lui répond.
Labour efficiency 2005-08-23
Rob pose la question :

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?


Harley Weston lui répond.
 
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