Surface Area = 4,536.45979 m2
The volume of every sphere is 4/3(pi) x (radius)3 .So the volume of your particular sphere is 4/3(pi) x (19)3.
Area = Base Length * Height So Base Length = Area/Height = 190 sq m/19 m = 10 metres.
1 meter = 100 centimeters100 cm divided by 19 = 5.26 pencilsTherefore, if a pencil is 19 cm long it will take 5.26 of them to make 99.94cm (as close to a metre as you will get).
The lateral surface area is A = pi*(radius)*(slant height), where (slant height) = sqrt(r^2 + h^2). So A = pi*(5 in)*sqrt((5 in)2+(19 in)2) = 308.6125 square inches
A rectangle can't have three dimensions. The object with those dimensions would have two surfaces 4 x 19, two surfaces 4 x 34 and two surfaces 19 x 34. 152 + 272 + 1292 = 1716 square units
First find the radius by dividing 19 by 2*pi which is 3.023943919 cm Surface area of a sphere (the ball) = 4*pi*radius2 Surface area = 4*pi*3.0239439192 => 114.9098689 Surface area of the ball: 115 square centimeters correct to three significant figures.
The surface area of a cone if the height is 21 and the diameter is 19 is 971.43 units2
Surface Area = ~791.68 ft2
Total surface area = (2*pi*112)+(22*pi*19) = 660*pi square inches
The ratio is 19/9.
The volume of every sphere is 4/3(pi) x (radius)3 .So the volume of your particular sphere is 4/3(pi) x (19)3.
o.19 m. 100 cm to a meter
Area = Base Length * Height So Base Length = Area/Height = 190 sq m/19 m = 10 metres.
There is no rectangular prism below 8 8 19.
19 inches is 0.48 meter.
V = 28,730 ft3
Weight = Volume times Density To answer this question the density of the rod has to be known, probably in Kg per Cubic Meter ( kg/m3) Volume of rod is Cross-sectional Area times Length Area for Square section rod is 19/1000 times 19/1000 = 0.000361 square meters Length is 1 meter Therefore volume is 0.000361 cubic meters Area for Round rod section is π*D squared / 4 or 22/7 * 19/1000*19/1000 / 4 = 0.000284 square meters. Length is 1 meter therefor Volume is 0.000284 cubic meters Weight is Volume times Density All units have to be compatible!