Would I use the Distance formula for this equation?
817425
(a) By drawing the points on a graph sheet, Midpoint is determined as (11,16)\n\n(b) Again by drawing on a graph sheet, by using Hypotenuse theorem; \nRadius = Longest side of right angled triangle \n            = square root of (4^2+5^2) \n            = square root of (16+25) \n            = square root of (41) \n            = 41^0.5 \n            = 6.4
603804
apply formula for it\nits easy
106436
I followed bugzpodder's advice.  Here's the formula I came up with. \n\n - 778728 - 121373345554191121 / 217945728000 * n^4 + 1298728131310487 / 1009008000 * n^3 + 670835515657 / 360360 * n + 2210058967013 / 12830400 * n^5 - 113499324114761 / 2874009600 * n^6 + 8961501223669 / 1306368000 * n^7 - 66868104743317 / 73156608000 * n^8 + 85853565361 / 914457600 * n^9 - 121097474346419 / 60540480 * n^2 - 27139677827 / 3657830400 * n^10 + 1792199 / 4009500 * n^11 - 1158076427 / 57480192000 * n^12 + 12283303 / 18681062400 * n^13 - 7672127 / 523069747200 * n^14 + 261523 / 1307674368000 * n^15 - 13147 / 10461394944000 * n^16\n\nCheers
17917
let me give you the distance formula: the square root of ( [X2 - X1]squared + [Y2 - Y1]squared )\n\nwhat are X2 and Y1 and stuff like that?  well, your coordinates are (-7,6) and (-1,1).  You assign one set of coordinates (it doesn't matter wich) to be x1 and y1.  the other set will be x2 and y2.  So, for example, in the point (-7,6) your -7 would be x1 and your 6 would be y1.  then your -1 would be x2 and your - would be y2.  make sense?  so once you have assigned your x1, x2, y1, and y2, then you just plug them into the distance formula:\n\nthe square root of ( [-1 - -7]squared + [1-6]squared ).\n\nyou solve that, and there's your answer!  I would use a scientific calculator to do the squaring and square rooting, and I assume you know which buttons do that.
200117
distance= rate*time\ntime= distance/rate
768807
distance formula\ndistance between two points =Square Root of ( (x1-x2)^2 + (y1-y2)^2)\nSquare root of ((4-1)^2 + (-8 -3)^2)\nSquare root of (9 + 121) \nSquare root of (130)\nuse calculator\n11.4
843247
You have enough information.\n\nIf I understand you correctly, you have the (N,E) coordinates for points A and F.  So given what you know so far, you should be able to compute coordinates for points B, C, and E.  All you need for a complete picture is coordinates for D.  Lets call them (N(D), E(D)).\n\nNow you know the distance DE. Using the distance formula,\n\n(DE)^2 = [N(E) - N(D)]^2 + [E(E) - E(D)]^2 \nthis is one equation with two unknowns.\n\nAlso, you know the azimuth for CD.  Call it θ.\nThen tan(θ) =   [E(D) - E(C)] / [N(D) - N(C)]\nthis is a second equation with the same two unknowns.\n\nTwo equations, two unknowns.  Solve simultaneously to get\nN(D) and E(D).\n\nOnce you have the coordinates you can compute missing azimuth and missing distance.\nIf you would like to email me the numbers I'll take a look.
871595
Let D = distance between home and school.\n\nUsing the formula : distance = speed * time, we have :\n\nFrom home to school : D = x * T1\nFrom school to home : D = kx * T2\n\nTherefore, x * T1 = kx * T2\nDividing through by x gives :\nT1 = k * T2\n\nWe know that T1 + T2 = 1 (total time travelled)\nTherefore, T2 = 1 - T1.\n\nSubstituting this into T1 = k * T2 gives :\nT1 = k (1 - T1)\n\nFrom this, we find that :\nT1 = k / (k + 1)\n\nSubstituting into 'home to school equation' gives :\nD = x * k / (k + 1)\n\nTherefore total distance covered\n= 2D\n= 2 * x * k / (k + 1) miles.
1114869
Here's a formula on this webpage you can use
34052
I do not know a preprogrammed utility to do this, but can point you in the direction that you might need to go.  See the reference below for the math and an on-line converter:
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
1088640
edit: (sorry hehe)\nthey'd be 500 miles apart...\n\nuse the pythagorean theorem...\n\nat 3 pm, one would be 300 miles west and the other would be 400 miles north...\n\nto get the distance between them (x):\n\nx=square root of 300*300 + 400*400\nx=500
119218
For distance, just add the magnitudes of all distances the object travelled.\n\nfor displacement, us the formula\n\nd = sqrt ( d1^2+d2^2+2*d1*d2*cos(angle between d1 and d2) )
243814
although the equation is not complete and therefore the full answer can not be given. i believe this equation will need to be differentiated using the product of a product rule.
538371
http://en.wikipedia.org/wiki/Cubic_equation
164104
maybe :\n\nKa = 1.3*10^-2