New Hampshire's coordinates are approximately 43.1939° N latitude and 71.5724° W longitude.
The country code and area code of Malindi, Kenya is 254, (0)42.
The minimum amount of paper Trevor needed to cover the entire cylindrical can is the lateral surface area of the cylinder. The lateral surface area of a cylinder can be calculated using the formula 2πrh, where r is the radius (5 cm) and h is the height (18 cm) of the cylinder. So, the minimum amount of paper needed is approximately 565.49 square centimeters.
The GCF of 18 and 42 is 6. 6 goes into 18 3 times and into 42 7timesTo find the GCF, you can list the factors of each number:The factors of 18 are 1, 2, 3, 6, 9, and 18The factors of 42 are 1, 2, 3, 6, 7, 14, 21, and 48The greatest factor they have in common is 6, so 6 is theGCF.The GCF is 6.
Latitude: 42°N to 46°18′NLongitude: 116°28′W to 124°38′W
Assuming that it's a rectangle then:- Area = 42*14 = 588 square cm Perimeter = 42+42+14+14 = 112 cm
Area 42 cm2, perimeter 26 cm.
28
Area of circle = pi*212 = 1385.44236 square cm Area of rectangle = 1385.44236 square cm Lenght of rectangle = 1385.44236/18 = 76.96902 cm Perimeter of rectangle = 2(76.96902)+2(18) = 189.93804 cm
The area is about 127.3 square inches.
Perimeter = 168 inches so 4*length of side = 168 inches Length of side = 168/4 = 42 inches So area = 42*42 = 1764 square inches.
That depends how you define the perimeter for a cube. The term perimeter is usually used for plane figures, not for 3D solids.
Even if you knew how many sides the polygon has, you stillcould not calculate its perimeter with that much information.Examples:-- An equilateral triangle with area of 20 has perimeter of 20.3885 .-- A square with area of 20 has perimeter of 17.889(rounded).-- A rectangle with area of 20 can have any perimeter more than 17.889 .4 by 5 . . . . area = 20, perimeter = 182 by 10 . . . area = 20, perimeter = 241 by 20 . . . area = 20, perimeter = 42..etc.
Perimeter is the distance AROUND any particular object where area is the amount of space inside the object (2dimentional). Let's say you have a yard 10 feet long, by 11 feet wide. The perimeter is the distance AROUND your yard or 10+10+11+11 = 42ft The AREA is the amount of space inside or L*W or 10*11 = 110ft^2 Suppose I change the dimensions of the yard without changing its perimeter (42 ft) and see what happens to the area. Suppose the length and width are 20ft and 1 ft respectively. The perimeter is still 20+1+20+1=42. The area however is 20X1 =20 ft2 Now if we take the yard to be a square one of 10.5 ft side then the perimeter still remains 42 ft. but the area becomes 110.25 ft2. Now suppose I assume that the perimeter of a circle is 42 ft. What will the area of such a circle be? Now we know that the circumference of a circle is 2x3.14(pi)Xr = 42 ft. Therefore r = 6.687 ft. r2= 44.715 ft2 therefore area of the circle is 4xpixr2= 561.63 ft 2. Thus we can deduce that given a known perimeter a circular shape has or occupies the largest area. What happens if the shape is irregular? Say a irregular polygon? Would my deduction that the cicrular shape has the largest area remain valid?
Largest = 86, Smallest 26
When the linear dimensions of a plane figure are quadrupled, its perimeter is quadrupled, and its area is multiplied by 42 = 16 .
The perimeter of a rectangle is 42. Meters. The length of the rectangle is threemeter less than twice the width.Mar