find the exact distance between the points (3,0) and (0, -sqr3?
97125
use the distance formula\nsqrt(3^2+(-sqr3)^2)=?
525749
The numbers you have are two points on a coordinate grid( the one with the x axis and the y axis. If you have some graph paper you could find the points. The first number in each set is the x-number and the second is the y number. You are trying to find the distance between those points. If you drew the dots on graph paper you could make a triangle out of them. One side is the x side and it is 3 units long(13 -10).  One side of the triangle is the y side and its 4 units long(20-16). Connecting those two points makes the third side or hypotenuse of the triangle. \nSo we use the Pythagorean theorem to find the distance.\nSo its 3squared + 4 squared =9 + 16 = 25.We must take the square root of 25 for the final answer, which is 5. Hope this helps.
811205
This is just an 8 by 7 rectangle, for an area of 56.\n\nYou could plot your points out to see the dimensions were 8 by 7.\n\nOr you could use the distance formula to get the distance between each point:\n\nD = √ [ (y2-y1) ² + (x2-x1) ² ]\n\nSo, for the distance between (-3,1) and (5,1) you have:\n\nD = √ [ (1-1) ² + (5-(-3)) ² ] = \n√ [ 0 + 8 ² ] = 8\n\nDo this for your other three pairs and you'd see that you had two distances of 8 and two distances of 7, for a regular 8 by 7 rectangle.\n\n~ ♥ ~
903023
M (-7, -3) is the midpoint of segment PQ and the coordinates of Q are (6, -3).\nLet the co-ordinates of P be (x,y).\nThen, (-7,-3) = ((x+6)/2, (y-3)/2)\n=>(x+6)/2 => -7 , (y-3)/2 = -3\n=>(x+6) => -14 , (y-3) = -6\n=>x => -14-6 , y = -6+3\n=>x => -20 , y = -3\n\nSo, co-ordinates of point p are (-20, -3)\n\nAll the best
490567
=-1\n2(0)+3(0)-1=f(x)\n2(0)=0\n3(0)=0\nminus 1\nequals negative 1 or -1
864521
y = 3/(1 - x)²\n\n1, (1 - x)² ≥ 0 for all x and therefore 3/(1 - x)² ≥ 0 for all x\n\n2. 3/(1 - x)² is undefined when x = 0. Therefore 3/(1 - x)² (ie y) → +∞ when x → 1 from the left of 1 and when x → 1 from the right of 1\n\n3. As x → ±∞, 3/(1 - x)² (ie y) → 0\n\n4. Plot a few other points and Bob's your uncle!!
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.
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
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(x-3)² / 25 + (y+2)² / 169 = 1\n\nie (x-3)² / 5² + (y+2)² / 13² = 1\n\nThus centre ≡ (3, -2)\n\nLength of semi-major axis (a) = 13 on vertical axis\nLength of semi-minor axis (b) = 5 on horizontal axis\n\nEccentricity (e) = √(1 - (b/a)²)\n= √(1 - (5/13)²)\n= 12/13\n\nFocal length = ea\n= 12/13 * 13\n= 12\nSo distance between the foci = 24\n\nThus focus 1 ≡(3, -2+ 12) ≡(3, 10)\nand focus 2 ≡(3, -2 - 12) ≡(3, -14)
249745
the answer is \n\n(sq.rt 3)(sq.rt 3) + 3(sq.rt 3) - (sq.rt 3) - 3\n\n= 3 + 2(sq.rt 3) - 3\n\n= 2(sq.rt 3)\n\nand that is worth 10 marks not 2
626632
The Distance Formula\nThe Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry. Here's how we get from the one to the other:\n\n \n\nSuppose you're given the two points (–2, 1) and (1, 5), and they want you to find out how far apart they are. The points look like this:\n\n\n  \n\n\n plotted points\n\n  \n\n \n\nYou can draw in the lines that form a right-angled triangle, using these points as two of the corners:\n\n\n  \n\n\n right triangle\n\n  \n\n \n\nIt's easy to find the lengths of the horizontal and vertical sides of the right triangle: just subtract the x-values and the y-values:\n\n\n  \n\n\n lengths\n   \n\nThen use the Pythagorean Theorem to find the length of the third side (which is the hypoteneuse of the right triangle):\n\n      c2 = a2 + b2\n\n...so: Copyright © Elizabeth Stapel 2000-2006 All Rights Reserved\n\n      c = 5\n\nThis format always holds true. Given two points, you can always plot them, draw the right triangle, and then find the length of the hypoteneuse. The length of the hypoteneuse is the distance between the two points. Since this format always works, it can be turned into a formula:\n\n      Distance Formula:\n\n      Given the two points (x1, y1) and (x2, y2), the distance between these points is given by the formula:\n\n            Distance Formula\n\nDon't let the subscripts scare you. They only indicate that there is a "first" point and a "second" point; that is, that you have two points. Whichever one you call "first" or "second" is up to you.\n\n    * Find the distance between the points (–2, –3) and (–4, 4).\n\n      Just plug them in to the Distance Formula:\n\n            d = sqrt(53)\n\n      Then the distance is sqrt(53), or about 7.28, rounded to two decimal places.\n\nThe commonest mistake made when using the Formula is to accidentally mismatch the x-values and y-values. Don't subtract an x from a y, or vice versa; make sure you've paired the numbers properly.\n\nAlso, don't get careless with the square-root symbol. If you get in the habit of omitting the square root and then "remembering" to put it back in when you check your answers in the back of the book, then you'll forget the square root on the test, and you'll miss easy points.\n\nYou also don't want to be careless with the squaring inside the Formula. Remember that you simplify inside the parentheses before you square, not after, and remember that the square in on everything inside the parentheses, including the minus sign, so the square of a negative is a positive.\n\nIn other words, if you do each step completely, instead of sloppily or in your head, then you're much more likely to get the right answers.\n\nBy the way, it is almost always better to leave the answer in "exact" form (the square root "sqrt(53)" above). Rounding is usually reserved for the last step of word problems. If you're not sure which format is preferred, do both, like this:\n\n      d = sqrt(53), or about 7.28\n\nNow YOU try it!\n\nVery often you will encounter the Distance Formula in veiled forms. That is, the problem will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember the Formula. For instance:\n\n    * Find the radius of a circle, given that the center is at (2, –3) and the point (–1, –2) lies on the circle.\n\n      The radius is the distance between the center and any point on the circle, so find the distance:\n\n            d = sqrt[ (2 + 1)^2 + (-3 + 2)^2 ] = sqrt(10)\n\n      Then the radius is sqrt(10), or about 3.16, rounded to two decimal places.\n\n    * Find all points (4, y) that are 10 units from the point (–2, –1).\n\n      I'll plug the two points and the distance into the Distance Formula:\n\n            10 = sqrt[(-2 - 4)^2 + (-1 - y)^2] = [y^2 + 2y + 37]\n\n
780311
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
140211
I am guessing you wanted to ask sqrt(8 exp(8)) . If this is the question then put the exact string in google and search.
692007
its quite simple the radius is half way threw the circle so its the radius 3 squared which equals 9 and 9 times pi or 3.14 equals 88.7364 but u mite have to round it so if u round to a whole number it would be 89 but to the nearest tenth it would be 88.7
791605
And find the slope of the line, and extend a line of the same slope and length from R to find the other endpoint.\n\na) slope = (-12 - 22)/(52 -12) = -30/40 = -3/4. Extending a line with the same slope and legth in the opposite direction can be done by going -40 units along x and 30 units along y. The other endpoint will then be (12 - 40, 22 -(-30)) or (-28, 52).\nb) For this we find the distance between the two endpoints. D = sqrt((52-(-28))^2 + (-12-52^2)) = sqrt(80^2 + 60^2) = sqrt(6400 + 3600)  = sqrt(100^2) = 10. The circle has a diameter of 10.
465898
sqrt(20/3)\n(sqrt(20))/(sqrt(3))\nMultiply top and bottom by sqrt(3)\n(sqrt(60))/3\n(sqrt(4 * 15))/3\n(2sqrt(15))/3
735493
The distance between two points is defined by the following equation derived from the pythagorean theorum:\n\nd = sqrt((x2 - x1)^2 + (y2-y1)^2)\n\nWhere (x1, y1) is the coordinates of the first point and (x2, y2) is the coordinates of the second point.\n\nFor your problem,\n\n(x1, y1) = (4, -3)\n(x2, y2) = (-4, 3)\n\nd = sqrt((-4 - 4)^2 + (3 - -3)^2)\nd = sqrt((-8)^2 + (6)^2)\nd = sqrt(64 + 36)\nd = sqrt(100)\nd = 10\n\nThe distance between the two points is 10.\n\nI hope this helps!