tHIS IS MATH  1.R=17ft 2.r=25.9 m 3.D=23.5 cm 4.d=82.1 mm 5.d=28 ft and 6.d=3.21 m?
492563
This cannot be the whole question.  Since you are using the letters r and d, I will assume you are talking about radius and diameter.\nAre you trying to find the area? then use the formula\nA = Pi*r^2\nIf you are trying to find circumference, then use\nC = 2*Pi*r\nRemember that d = 2r (diameter is twice the radius)
869579
let the height of cliff be h\nthen \ndistance of first ship =x\ndistance of first ship =x+500\nnow \n1.h/x=tan 24\n2.h/(x+500)=tan 15\nfrom 1\nx=hcot24\nfrom 2\nx=hcot15-500\ntherefore\nh(cot15-cot24)=500\nh=336.47
120237
L and W are Length and Width, respectively,\n           2L + 2W = 210 ----- (i)\n           L - 2W = 15 ------(ii)\n Solving Equation (i),\n           2L + 2 W = 210\n         2 ( L + W )= 210\n           ( L + W )= 105  \n                  L = 105 - W\nSolving Equation (ii)\n            L - 2W = 15 \n                 L = 15 + 2 W\nEquating both the expressions,\n             105 - W = 15 + 2 W\n             105 - 15 = 3 W\n                   90 = 3 W\n                   30 = W\nPutting the Value of W in equation (i),\n                 2L + 60 = 210\n                      2L = 210 - 60\n                      2L = 150\n                       L = 75\nTherefore, the length is 75ft and the width is 30ft
734759
v= 4 pi r^3 /3\n\ndv/dt = 4 pi r^2  dr/dt\n\n5=   dr/dt\n\ndr/dt= 5/(4 pi r^2)\n\ndr/dt = 5/(4 pi 3^2) = 5/36 pi\n\ndiameter = x = 2r\n\ndx/dt = 2 dr/dt\n\ndx/dt = 5/36 pi* 2 = 5/18 pi
839453
1) 21.69\n2) 14.10\n3) 7.59
398633
Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
30686
Let the length of the equal ropes be x ft\n    the length of the longer rope is (x+2)ft\n\ntherefore,\n\nx+x+x+(x+2) = 44\n4x = 44-2\n4x = 42\nx = 10.5
812431
2.5 m= - 25.. are you trying to solve for M.... \nand -3.25t=-3.11 i am assuming you are trying to solve for T?\n\nfor the first one... 2.5m/2.5=-25/2.5 so m = 10\n\n2nd one .. -3.25t/-3.25= -3.11/-3.25\n\ni don't know if this is what u looking for
774792
Let the length of side "b" be x ft.\nTherefore side a=2x and side c=2x+3\nAccording to the problem,\nx+2x+2x+3=93\n=>5x+3=93\n=>5x=93-3=90\n=>x=90/5=18\nTherefore.side a=2X18=36 and side c=36+3=39\nAnswer a=36ft. b=18 ft and c=39 ft
927311
First of all, you don't have to use the Pythagorean theorem for this. Find x, so 6x=54. x=9. Then, divide 45 by 9, because the lines are congruent, and you get 5 ft.
362534
1 FT = 12 INCH\n1 INCH = 2.54 CM\n\nTHERE FORE:\n\n5 FT + 6 INCH = 66 INCHES\n66 INCHES = 167.64 CM
475727
the answer to the first question is 5 m/s.\nF=ma; a=F/m=200N/40kg=5m/s\nthe second answer is 1m/s.\nF=ma; a=F/m=2N/2kg=1m/s\nthe third answer is 1:4.\nF=ma; m=F/a\n(1) M=f/a=F/a=m\n(2) M=f/a=4F/a=4m
364710
3 feet is 36 inches.\n\n0.8 feet is 8/10 of a foot. Solve 8/10 = x/12 for x, and you get 9.6 inches.\n\nSo 3.8 feet is 36 + 9.6 inches, or 45.6 inches.
1073112
First convert everything to feet. The perimeter of 25 1/3 yards x 3 = 76 feet.\n\nThe length of the house is 22 feet. So the two lengths add up to 44 feet of the perimeter. That leaves\n\n76 - 44 = 32 feet for the two widths. So one of those widths would be:\n\n32 / 2 = 16 feet.
822454
1.  = 1cm\n2.  = 2cm\n3.  = 4cm\n4.  = 7cm\n5.  = 11cm\n6.  = 16cm\n7.  = 22cm\n8.  = 29cm\n\ntotal length = 92cm\neach step adds more to the next length.  \nbetween the first and second step, add 1\nbetween the second and third step, add 2\nbetween the third and fourth step, add 3\nbetween the fourth and fifth step, add 4....\n\nand so on.
1134148
Since sheriefhalawa didn't give you an answer, but merely told you how to find it, I expect this will be of more interest:\n\nImpulse(1) = 26 * .056 = 1.456 Ns\nDv(1) = I(1)/m  =  9.1 m/s\nv(1) = 3.1 + 9.1  =  12.2 m/s\n\nI(2) = -26*.056 = -.6608 Ns\nDv(2) = -.6608/.16 = -4.13 m/s\nv(2) = 3.1 - 4.13 = -1.03 m/s
930657
So they told you the relation is:\nd=kh\n(Since they said "varies directly", there is no offset: d=kh+d0)\nThen: 8.45=6.5k\ntherefore k=1.3 (unitless, ft/ft)\n\nThen you want to find:\nd=5.2 = kh=1.3h\nTherefore:\nh = d/k = 5.2/1.3 = 4 ft