Calculus Word Problem?
595452
If you picture your trapezoid with the shorter parallel side and the two legs as sides 4, then the only unknown is the the long parallel side.  That side is at least 4, but with two extra parts on either side (if you drop perpendiculars from the short to the long side, you'd have a rectangle and two triangles).  So the long side is 4 + 2x, where x is the length of those two parts, which is also a leg of a triangle that we created in dropping those perpendiculars.\n\nThe height of the trapezoid is the same as the other leg of the triangle.  We don't know the length of either leg, but we do know that the hypotenuse is 4.  So through the Pythagorean theorem, we have\n\n4^2 = h^2 + x^2\nh^2 = 16 - x^2\nh = sqrt(16 - x^2)\n\nBack to the trapezoid - we know the area of a trapezoid is the average of the bases times the height, so we have\n\nA = (1/2)(4 + 4 + 2x)(sqrt(16 - x^2))\n= (4 + x)(sqrt(16 - x^2))\n= (4 + x)(16 - x^2)^(1/2)\n\nTo find the maximum area, take a derivative, set it to 0, and solve for x:\n\nA' = (4 + x)[(1/2)(16 - x^2)^(-1/2)](-2x) + [(16 - x^2)^(1/2)](1)\n= (-4x - x^2)(16 - x^2)^(-1/2) + (16 - x^2)^(1/2)\n\nIf we set this to 0, then\n0 = (-4x - x^2)(16 - x^2)^(-1/2) + (16 - x^2)^(1/2)\n-(-4x - x^2)(16 - x^2)^(-1/2) = (16 - x^2)^(1/2)\n4x + x^2 = 16 - x^2\n0 = 2x^2 + 4x - 16\n0 = x^2 + 2x - 8\n0 = (x + 4)(x - 2)\nx = -4 or 2\n\nSo your long side has to be 4 + 2(2) = 8 to maximize the area.
118732
good luck man... http://mathforum.org/library/drmath/sets/high_calculus.html
244823
The amount (A) will be doubling, so substitute "2P" in for A. Then you can drop the "P" from each side by division and solve for t.\n\nA = Pe^(rt)\n2P = Pe^(rt)\n2 = e^(rt)\nln 2 = (rt) ln e\nln 2 = .054t (ln e = 1)\n(ln 2)/.054 = t\n(approx.) 12.836 = t\n\nDoes that sound correct? (It's been awhile since I've done these kinds of problems.)
808176
Correct answer should be E
924613
http://www.mathwords.com/\n\n\nexcellent site.
977084
a.) newtons\nb.) km/h\nc.) f(x) = 0\nd.) x = f(0) + c
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
741552
You have got it right.\nthe answer is\n\nf(x) = 3x
1130821
so for the first part, f(x)=x \nthe derivative no matter what is 1 \n\nfor the second part f(x)=1\nthe derivative no matter what is 0 \n\nso you will have combinational solution
2024
Case 1: f(x) := x-5*y; (y is parameter)\n\nL{f(x)} = int(exp(-s*x)*f(x),x=0..infinity);\nL{f(x)} = int(exp(-s*x)*(x-5*y),x=0..infinity);\nL{f(x)} = 1/s^2-5*y/s;\n\nCase 2: f(y) := x-5*y; (x is parameter)\n\nL{f(y)} = int(exp(-s*y)*f(y),y=0..infinity);\nL{f(y)} = int(exp(-s*y)*(x-5*y),y=0..infinity);\nL{f(y)} = x/s-5/s^2;
589798
y' = 2x - 2\n2x - 2 = 0\n2x = 2\nx = 1
500014
a. (x+9)(x-9)\nb.(t-2)(t+2)\nc. (b-12)(b+12)
914435
x > 5\n\n- - - - - -s-
625162
Try dictionary.com, m-w.com, or one of the thousand other good dictionary websites on the web, as for problems, try search on http://mathforum.org/
276361
Goto this site and look at problem #24 and you will see the solution to this problem...in the end the answer sec x.\n\nhttp://noether.uoregon.edu/~montgome/math251/homework/Section3.8.pdf\n\nI hope this helps...\n\nDon
912865
bring the a's to one side... u then get 9a = 27 ... divide and answer is a = 3
609300
This is more of a word problem or a riddle than a math problem.\n\nThe answer is 7 (SEVEN).  Remove its first letter and it becomes EVEN!