Please help this is the last few questions
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Answers: 2
Mathematics, 21.06.2019 16:30
Refer to the table below if needed. second quadrant third quadrant fourth quadrant sin(1800- - cos(180° -) tan(180°-e) =- tane cot(1800-0) 10 it to solo 888 sin(180° +c) = - sine cos(180° +) =- cose tan(180° +c) = tane cot(180° +o) = cote sec(180° + c) = - seco csc(180° +2) = - csce sin(360° -) =- sine cos(360° -) = cose tan(360° - e) =- tane cot(360° -) = -cote sec(360° -) = seco csc(360° -) = csco sec(180° -) = csc(180° -) = csca 1991 given that sine = 3/5 and lies in quadrant ii, find the following value. tane
Answers: 2
Mathematics, 22.06.2019 01:00
For every corresponding pair of cross sections, the area of the cross section of a sphere with radius r is equal to the area of the cross section of a cylinder with radius and height 2r minus the volume of two cones, each with a radius and height of r. a cross section of the sphere is and a cross section of the cylinder minus the cones, taken parallel to the base of cylinder, is the volume of the cylinder with radius r and height 2r is and the volume of each cone with radius r and height r is 1/3 pie r^3. so the volume of the cylinder minus the two cones is therefore, the volume of the cylinder is 4/3pie r^3 by cavalieri's principle. (fill in options are: r/2- r- 2r- an annulus- a circle -1/3pier^3- 2/3pier^3- 4/3pier^3- 5/3pier^3- 2pier^3- 4pier^3)
Answers: 3
Mathematics, 22.06.2019 03:00
Solve 2x − 1 = 11 for x using the change of base formula log base b of y equals log y over log b.
Answers: 3
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