application problem[math]?
953146
Let the number of locals be x,  and LD calls y.\n\nParental limits are\nx <= 30  and  y ......\n\nThe total time spent is 10x + 30y minutes, so you can write an inequality representing a need to spend at least 240 minutes on the phone.  I won't do that for you, but probably someone else already has!\n\nNo!  I see they haven't yet. I'm sure you can do it.
743387
Working above is good until this step:\n\n2 + 0.8x >0.9(x + 6)\n2 + 0.8x >0.9x + 5.4)\n0.1x > -3.4 ← Symbol error in this line\n\nShould read 0.1x < -3.4\nie x < -34
137681
Ok...\nI see what you are doing...\nYou merely substitute the number in for x...\nso the problem becomes 6-2*(-1)\n6+2\n8
805339
These are quite famous problems. As you mentioned, you can do it with integration; thats probably the simplest way.\n \nIn order to see why that works, lets call f(x) = 1/x or 1/x^2 depending on which one you want. \nDraw a 'histogram', where each of the bars is 1 wide, and has height 1/(left x value). So you have a bar from x=1 to x=2 with height 1, x=2 to x=3 with height 1/2, etc.\n\nNow if you draw the curve f(x), it never goes below any of these bars. So the integral of f(x), which is the area under the curve, must be less than the total area of each bar, which is that infinite sum. So if we get a finite integral, we have must a finite sum.\n\nSimilarly, you could draw a histogram where each bar has height 1/(right x value). Then the curve is always *below* the bars, so you get a lower limit. Then if that integral is infinite, the area must be infinite.\n\nAnyway, now that we've got the official answer sorted, there are many other proofs. Heres a nice one for the 1/k case:\n= 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ...\n>= 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + ...\n= 1 + 1/2 + 1/2 + 1/2 + ...\nwhich goes to infinity.\n\nAs a side note, the actual sum of 1/k^2 is pi^2/6. That always interests me - if you draw a whole bunch of circles with radius 1, 1/2, 1/3, .., then their total perimeter turns into the 1/k-type sum, and is infinite. However, their total area turns into the 1/k^2-type sum, and is finite. So you can't draw the outline of the circles, but you can colour them in.. ;)
208332
What problem are you having? You did not provide any code, or even a possible algorithm. Try actually providing us with some better information, so that we can actually help you.
199081
Sorry, but I think you need to do this yourself.  Here are some guidelines though; give them a try.\n1.  Make a line drawing so you can see the info you're given and the path to the required answer.\n2.  Make a list of ALL the related formulae you were given.  Check each one against the given data; if you find one in which you have all the variables EXCEPT one, substitute and solve for the unknown.\n3.  Sometimes, you may have to use one formula to find a variable and THEN use that in ANOTHER formula to arrive at your ansver.\n\nGood luck!!
424650
Not really enough information to help, so try this site:\n\nhttp://www.blc.edu/fac/rbuelow/calc/ex4-6.html\n\nit's the second problem down the page
1185586
this has happened to me before on my 89. it's usually when the expression is too complicated--that is, it would take the calculator more memory than it has to solve it. so i'd recommend trying to first simplify the expression and then enter it. For instance, don't leave things in terms like sin(2) when you can just get an approximate number for that. maybe that's it?
448711
51,336\n\nNext time just type your math problem into at google search and it will do the math for you!\n\nLink below.
735282
Seems like you are missing some parentheses and an exponent. Please check to make sure yiu have copied the problem correctly and resubmit.
616629
First of all, change this:\n\n<p>Enter first numerator: <input type=text name=num1 size=20></p>\n<p>Enter first numerator: <input type=text name=num2 size=20></p>\n<p>Enter second denomenator: <input type=text name=num1 size=20></p>\n<p>Enter second denomenator: <input type=text name=num2 size=20></p>\n\n\nto this:\n\n<p>Enter first numerator: <input type=text name=num1 size=20></p>\n<p>Enter first denomenator: <input type=text name=num2 size=20></p>\n<p>Enter second numerator: <input type=text name=num3 size=20></p>\n<p>Enter second denomenator: <input type=text name=num4 size=20></p>\n\nThen based on the operation selected, implement the proper logic there (its basic math) and then your program will work.\n\nGood luck!!
893957
well darling ;\nimagine this like that\n\ny^2 + ( a+b)y + ab \nyou need to find " a+b " & " ab" \nin our function \na+b = 16 & ab = 64 \n( y + a)( y + b) \n( y + 8 ) ( y + 8)\n{ ' 8 + 8 = 16 '   &  ' 8 * 8 = 64 ' }\nso\n( y + 8)^2\n\nGood Luck sweetheart.
708619
Please state the requirements of the mathematic's work project.\n\n- - - - - - - -s-
1081089
[ (50000 x 2.4)+(25000 x .8) ] / (50000+250000)=1.8\n\nThe remainder of the info in the problem is not necessary.
1144014
draw a picture of the graph over the specified interval.  divide the graph into vertical rectangles.  the area under the curve will be the sum of the areas of the rectangles.  saying that youve made "n" rectangles, the only way to get the exact area is to find the sum for the limit as n approaches infinity.  work from that concept.  if you already knew that, or need more help or something, email me and i can give you more steps and a much more detailed description.  that would take a really long time though so im gonna see if this helps first.
1066144
Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.
820052
I do not understand how the "problem" has to do with the question. But to answer the first part...\n\nThe cell should contain \n=POWER(number, power)\n\neg. =POWER(5, 2) would display 25\n\nA cubic root would look like this...\n=POWER(number, 1/3)