how do you find the exact distance between the points:  (3,0) and (0,-sqr3)  ?
144367
The distance between two pints (a,b) and (c,d) is\n sqrt((d-b)^2+(c-a)^2). \nThen,\nd = sqrt ( (-sqr3-0)^2 + (0-3)^2) = sqrt ( 3 + 9) = sqrt (12) = sqrt (4*3) = 2 sqrt 3 = 2* 1.73 = 3.46
829059
I hope I did this correctly.  Would you let me know?\n\nI set up the equation of the relative distance the particles will have from eachother at time t:\n\ndistance equals location of particle one minus location of particle two:\n\nx=\n(4t-3)i+\n(6-3t)j+\n(5t)k\n\nthen I distributed the i, j and k to get\n4it-3j+\n6j-3tj+\n5tk\n\nThen I took the derivative dx/dt treating i, j, and k as constants to get\n4i-3j+5k\nthis is the minimum separation\n4i-3j+5k\n\nThis I then used to set up the following three simultaneous equations to find the time that occurs:\n\n4t-3=4\nt=1\n\n6-3t=-3\nt=1\n\n5t=5\nt=1\n\nj
1156233
The dist is\nsqrt.[(4-3)^2+(1-1)^2+(4-3)^2]=sqrt.2
137113
Draw the two points (a,b) and (c,d) on a graph so that the segment connecting them is NOT horizontal or vertical. Then make a right triangle with one vertical leg and one horizontal leg where the connecting segment is the hypotenuse.  \n\nThe length of the horizontal leg is c-a\nThe length of the vertical leg is d-b\n\nSo the length of the hypotenuse, which is the distance from (a,b) to (c,d), is\n\nsquare root of (c-a)^2+(d-b)^2
791994
First distance: Pythagorean Theorem is a squared + b squared = c squared.  So (2*2) + (3*3) = ???  Then find the square root of ??? \n\nThen the bearing: We first need to find the measure of angle A.  Since we know the length of all 3 sides, any of the SohCahToa formulas work.  Tangent is easiest here.  tan(A) = opposite/adjacent (length of opposite leg divided by length of adjacent leg).  tan(A) = 3/2.  Use the tan-1 key (probably will require shift key).  \n\nWe know the first leg of the trip is south which represents 180 degrees.  So add the found angle to 180 degrees.  (Draw it out, it should make sense.)
947787
In this equation, you have a right triangle.  After 3 hours, the first car has traveled 150 km south and the 2nd car ahs traveled 210 km west.  You can find the between these two cars at 3 hours by using the Pythagorean Theorem since you have a right triangle.\n\nDistance = sqrt(150^2+210^2) = 258.07 km\n\nThe general form of the Pythagorean Theorem is:\n\na^2 + b^2 = c^2, where:\n\na = distance of car 1\nb = distance of car 2\nc = distance between cars.\n\nTake the derivative to find how the rates are related to one another.  Take the derivative with respect to time:\n\n2a(da/dt) + 2b(db/dt) = 2c(dc/dt)\n\nIn this problem, we are looking for dc/dt.  These are our known values:\n\na = 150 km\nb = 210 km\nc = 258.07 km\nda/dt = 50 km/hr\ndb/dt = 70km/hr\n\nPlugging these values for the equation above, solve for dc/dt:\n\ndc/dt = [2a(da/dt) + 2b(db/dt)]/[2c] ------->\n\n[(2*150*50) + (2*210*70)]/(2*258.07)\n\nTherefore, dc/dt = 86.02 km/hr.  Now change to km/sec:\n\n(86.02 km / 1 hr)(1 hr / 60 min)(1 min / 60 sec) = 0.024 km/sec\n\n-----------\n\nHope this helps
329806
The distance from the center to a point on the sphere is the radius.  Find that by the distance formula, then double it to get the diameter.\n\n(The distance formula says to subtract the two x coordinates and square this difference.  Then do the same for the two y coordinates and z-coordinates.  Add these 3 answers and find the square root of the sum.  That's the distance.)
882750
Listen to the first guy.\nThe distance on the y axis is 1 ((-5)-(-4)) and on the x axis it is 3 ((-3)-0).\nThis gives you a right angles triangle with sides of 3 & 1. \nUnder Pythagoras theoem the sum of the squares is equal to the square of the hypoteneuse so\n1^2 + 3^2 = 10.\nTherefore the distance is the square root of 10.
329126
You have to graph them and find the corner points.  One of the corner points will be the answer.\nI found the corner points to be\n(28,52) and (40,40)\nPlug each point into g = 3x+y \nFor (28,52): g = 3(28) + 52 = 136\nFor (40,40): g = 3(40) + 40 = 160\n136 is the smaller value, so (28,52) is the answer.
35596
The answer above didn't go far enough down the page. That's just the equation of the line.\n\nIf the point is (x_0, y_0), then the distance is:\n\n(|ax_0+by_0+c|)/(sqrt(a^2+b^2)).
415223
distance to x-axis is √(8² + (-1)²) = √65.\ndistance to y-axis is √(3² + (-1)²) = √10.\ndistance to z-axis is √(3² + 8²) = √73.
211889
For distance just feed the formula:\nd =√((x1-x2)^2+(y1-y2)^2)....square root of the sum of the squares of the differences of the coordinates.  so subtract 2 and -3 to get 5, subtract -1 and 8 to get -9.  square the 5 and -9 to get 25 and 81, add to get 106, and take sqrt to get d=10.296\n\nFor the system of equations, go matrix/edit, pick matix [A], set dimensions to 2x3, enter coefficients 1, -1, 3 from the 1st equation into the 1st row of the matrix, and 2, 3, 11 from the 2nd equation into the 2nd row of the matrix.  quit, then matrix/math/B (rref( matrix/name [A].  rref is reduced row echelon form.  the answer looks like:\n[[1..0..4]\n.[0..1..1]]\n\nignore the dots, which are for spacing.  the left 2 columns are the 2x2 identity matrix.  the right column is the solution matrix:  x=4, y=1.
411757
since y=+-(x)^(3/2) Range [0,oo[\nnP=(n,n*n^1/2) and x_n/y_n= [(n^1/2^)/n]=n^-1/2
158873
the two points you have should each have an x coordinate and a y coordinate and should be in the form (x1,y1) for the first point and (x2,y2) for the second point\n\nto find the slope, divide y1-y2 by x1-x2 or y2-y1 divided by x2-x1...they will both give you the same answer\n\nto find the distance you will need to remember a little bit of trig.  remember that a squared plus b sqared equals c squared?  well a is going to be x1-x2....b is going to be y1-y2....and the distance of the line segment will be c....so (x1-x2)squared + (y1-y2)squared = the distance squared....so the distance is the sqaure root of [(x1-x2)sqared+(y1-y2)squared]\n\nto find the midpoint  take (x1+x2) and divide by 2 to get the x coordinate of the midpoint and (y1-y2) divided by 2\nto get the y coordinate for the midpoint\n\nin summary\nslope:  (y1-y2)/(x1-x2)\ndistance: [(x1-x2)^2+(y1-y2)^2]^(1/2)\nmidpoint (xm,ym): xm=(x1+x2)/2 and ym=(y1+y2)/2\n\nnote:  the ^ is what i'm using as the exponential operator...in other words something ^2 means it is to the second power.  if i did something ^3 it would be something to the third power....something^(1/2) means to the half power which is the same as square-rooted (you'll learn that in a later class) so:\n\n2^2=4\n3^2=9\n9^(1/2)=3\n16^(1/2)=4\n\nhope i didn't confuse you even more and remember you will really learn all of the stuff they are teaching now by doing homework for the next math class you will take. you will learn the stuff for that class in the next class.  the only way to really learn the stuff in the last math class you ever take is to teach it.
190089
use heron's formula: \nArea=SQRT(s(s-a)(s-b)(s-c)), \nwhere s=(a+b+c)/2 or perimeter/2. \n\nyou only need to know the sides of the triangles. \nfrom (0,9) to (4,6), the distnace is a=5\nfrom (0,9) to (-1,12) the distance is b=sqrt(10)~3.16\nfrom (4,6) to (-1,12) the distance is c=sqrt(61)~7.81\n\nso, s=(5+3.16+7.81)/2=7.985\n\nplugging these 4 numbers into the formula gives \nSQRT(7.985(7.985-5) (7.985-3.16) (7.985-7.81))=4.486\n\nso the area is approximately 4.486
1007342
y = -1 is a horizontal line that contains the points (n, -1)\n\nhow far is (3, 4) from that line?  Subtrace the y-coordinates: 4 - (-1) = 5
1079894
The basic definition of a parabola states that it is the locus of points that are equidistant from a given point (the focus), and a line (the directrix). Therefore, we have to find two distances:\n\nPoint-to-Point: d1 = sqrt( (x-11)^2 + (y+7)^2 )\nPoint-to-Line: d2 = abs( x-5 ) / sqrt (1^2) = (x - 5)\n\nThen, make the two distances equal, d1 = d2:\n(x-11)^2 + (y+7)^2 = (x-5)^2\n........simplifying........\ny^2 + 14y + 49 - 12x + 96 = 0\n\n(y + 7)^2 = 12*(x - 8)\n\nThis is the equation for our parabola.