In the figure, M and N are the midpoints of RT and ST. What is the length of MN?
402322
If M and N are midpoints of RT and ST, then MN is parallel to RS.  There is a property of triangles that says that if you have a base and a segment inside the triangle whose endpoints are midpoints of the other two sides, the segment is half the length of the base.\n\nSo in your case, we need to find the length of the base RS.  Since the points are on the x-axis, the length is simply |2a-2b|.  Thus, the length of the segment MN is (|2a-2b|)/2, or |a-b|.
136179
An inch is defined to be equal to the length of 2.54 cm.\n1 centimeter = 1/100 of a meter.  1 meter is defined to be The length of the path traveled by light in vacuum during a time interval of 1/299 792 458 of a second.  1 second is defined to be the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom.
244230
I do not clearly understand your question, but if you have a metric ruler and you want measure a drawing (or map) in English units scale, the problem is of conversions i.e. if drawing is 0.5=12" it would mean 1:24 so if you measure 10mm it would mean 240mm. Hope it helps.
865942
b = 70sin55\n\nb = 57.34
557508
T = 2 * pi * sqrt(l/g)\n2 = 2 * pi * sqrt(0.9942/g)\n--- taking your length to be in metres\n\ng = 9.8123607 m/s2
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
917169
L = μ(0) μ(r) N^2 A/l\n\nwhere\n \nμ(0) is the permeability of free space (4π × 10-7 henrys per metre) \nμ(r) is the relative permeability of the core (dimensionless) \nN is the number of turns. \nA is the cross sectional area of the coil in square metres. \nl is the length of the coil in metres\n\nHere we assume that μ(r) = 1, since the solenoid doesn't enclose anything like a metal core. Then\n\nL = [μ(0) N^2 (πR^2)] / l, where R is the radius of the solenoid\n\n= [(4π × 10-^7)(2920)^2(π(0.005)^2)] / (0.1)\n\n= 8.44 × 10-^3 henries
1118574
Find the midpoint of each interval.\nFor 1-3, the midpoint is (3+1)/2 = 4/2 = 2\nFor 4-6, the midpoint is (6+4)/2 = 10/2 = 5\nFor 7-9, the midpoint is (9+7)/2 = 16/2 = 8\n2(1) + 5(6) + 8(3)\n= 2+30+24\n= 56\nn = number of data points = 1+6+3 = 10\nvariance = [1/(10-1)][((2^2)(1) +(5^2)(6) +(8^2)(3)) - (56^2)/10]\n= (1/9)((4+150+192) - 3136/10)\n= (1/9)(346 - 313.6)\n= (1/9)(32.4)\n= 3.6\nstandard deviation = sqrt(variance)\n= sqrt(3.6) \n= 1.9
749000
To avoid any confusion I have called the point of intersection of MO and NP, Q\n\nΔMNP is isosceles (MN = MP)\n\nIn Δ's MNQ, MPQ\nMN = MP (Data)\n<NMQ = <PMQ (Data)\nMQ is common\nTherefore ΔMNQ ≡ ΔMPQ (SAS Property)\n\nTherefore NQ = PQ (Corresponding sides of congruent triangles are equal)\n\nAnd <MQN = < MQP (Corresponding angles of congruent triangles are equal)\nBUT <MQP is a straight angle\nTherefore <MQN = < MQP = 90°\n\nSo MO is the perpendicular bisector of PN
1010902
This is  basic\nslope= m=(y2-y1)/(x2-x1)=(12-7)/(8-6)=5/2\nangle=arctan(5/2)
756487
176.53 cm\n\nhttp://www.csgnetwork.com/csgcvtftin2cmeterscalc.html
602189
u can go www.webmath.com, its pretty good. \nu can also ask on answers. \ni.e. 1cm = 10mm\n1m = 100cm or 1m = 1000mm\n1m = 3ft\n1ft = 12in \netc... hope it helps!
653629
SI unit for length is 'meter'
249338
14400 inch\n36576 centimeter\n365760 millimeter\n400 yards\n0.3657611 kilometer\n365.76 meter
893495
T = F*d = I*alpha\n\nI = (1/3)*M*d^2\n\nIn both cases, d is the distance from the axis to "the outer edge of one pane",  "as shown in the drawing".
790030
Because the focal point is half the distance of the radius of curvature, the focal point is 4 cm.\n\n1/f = [1/d(image)]  + [(1/d(object)]\n\n1/4 = 1/17 + 1/d(image)\n\n1/d(image) = 1/4 - 1/17 = 17/68 - 4/68 = 13/68\n\nd(image) = 68/13 = 5.23 cm
136523
44.89Nm which is 44.9 Nm\n\nFormula is   torque=force*distance