Linear Algebra1882 - 68 من الصفحات |
طبعات أخرى - عرض جميع المقتطفات
عبارات ومصطلحات مألوفة
abstract number angle AOX conjugate consequently curvature curve diagonal ds ds ds ds2 equation f₂ Formula Formula 17 given point gonals hence hight II+II Let OA Let us suppose line MN line OA line which joins Linear Algebra linear unit multiplication Numerical Algebra OB=ß OC=Y OR+OS origin osculating plane parallel parallelogram pendicular plane which passes plane XOA principal direction OX Quaternions represent Rule of Signs side special perpendicular square ß ß ßß straight line Tait triangle velocity w₂ Y ₂ Y II Y₁ Y₂ α β α Π αα αβ αβ αβ βγ αγ αγ βα βγ βγ βδ γα γδ δβ Δρ Π Π ρβ ПП
مقاطع مشهورة
الصفحة 50 - Thus a parabola is the locus of a point which moves so that its distance from a fixed point is equal to its distance from a fixed straight line (see fig.
الصفحة 10 - The perpendicular bisectors of the sides of a triangle meet in a point which is equidistant from the vertices of the triangle. Let the -l. bisectors EE' and DD
الصفحة 23 - The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its four Dem.— Let ABCD be the a.
الصفحة 32 - To express the cosine of an angle of a spherical triangle in terms of the sines and cosines of the sides. Let ABC be a spherical triangle, O the centre of the sphere.
الصفحة 43 - The squares of the sides of any quadrilateral exceed the squares of the diagonals by four times the square of the line which joins the middle points of the diagonals. Retaining the figure and notation of Ex. 8, Art. 7, we have squares of sides as vectors and squares of diagonals therefore the former sum exceeds the latter by a2 + p° -t- y" - 2Sap - 2 A 2 2J Therefore as lines the same is true.
الصفحة 10 - AG2 = = line DA and similarly of the others. Ex. 7. The middle points of the lines which join the points of bisection of the opposite sides of a quadrilateral coincide, whether the four sides of the quadrilateral be in the same plane or not. Let...
الصفحة 9 - Prob. 11. If the perimeter of a right angled triangle be 720, and the perpendicular falling from the right angle on the hypothenuse be 144 ; what are the lengths of the sides I Prob.
الصفحة 46 - Consequently the condition that all three perpendiculars shall meet in a point is that the sum of the squares of each pair of opposite edges shall be the same.
الصفحة 24 - London. Three times the sum of the squares of the distances of the centres of the four circles touching the sides of a plane triangle, from the centre of the circumscribing circle, is...