should be general enough so that if the content of the input file changes and the program is run again(without editing and recompiling)‚ it outputs the appropriate results. Rectangle: Length = 10.20‚ width = 5.35‚ area = 54.57‚ parameter = 31.10 Circle: Radius = 15.60‚ area = 764.54‚ circumference = 98.02 Name: Randy Gill‚ age: 31 Beginning balance = $18500.00‚ interest rate = 3.50 Balance at the end of the month = $18553.96 The character that comes after A in the ASCII set is B
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identify each room‚ input the length (l) and width (w) of each room‚ multiply the l x w of each room to get the square foot‚ and add the sum of each room to secure the total square footage of the house. Room1 – Area = L x W Room2 – Area = L x W Room3 – Area = L x W Room4 – Area = L x W Room1 + Room2 + Room3 + Room4 = Total Square Footage III. Design Main Module Call Input Data module Call Perform Calculations module Call Output Results module End Program Main
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Michael Brause CMIS 102-6387 May 30‚ 2015 Program description: I calculated the usable area in square feet of a house. Assume that the house has a maximum of four rooms‚ and that each room is rectangular. I wrote pseudo code statements to declare 4 Integers and labeled them homesqft‚ room1‚ room2‚ room3‚ and room4. Each room will have its length and width to calculate its area. Analysis: Test Case # Input Expected Output 1 Room1: length=10‚ width=14 Room2: length=9‚ width=10 Room3: length=12
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V=π =π Given =- = x =π =-π(0.36 x =-0.39x1m When x=0.12 =π (0.12 Therefore:-0.39x1=π (0.12 =-7.5x1Therefore the water sinking rate is -7.5x1m QUESTION 3 : The radius of the sphere is r‚ the area of the sphere and it’s volume is π. If‚ when the r of the sphere is 21m‚ it is
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and as the river gets faster it carries greater amounts of bed-load” In my conclusion I will be linking all of these hypotheses to velocity of the river. Carding Mill valley (long Mynd): The name Long Mynd means "Long Mountain”‚ it is a rural area and it is one of twelve valleys the landscape is very hilly and steep and because of this there are not many buildings. (This can be seen on the contour lines).
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a copper wire is measured by using a system called the American Wire Gauge also known as AWG. This system has predominantly been used in the United States and Canada since 1857. Using this measurement the cross-sectional area of a gauge is able to be found. We then use this area to determine the wires current carrying capacity. The gauge numbers for wire and diameter values work inversely‚ as when one increases the other decreases. AWG tables are commonly used to portray values for a solid round
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Description This program will determine the usable area of a house. The rooms are rectangular in shape and maximum of four rooms are allowed to be calculated at once. It will use the area formula for a rectangle to find the usable area of each room and once the calculation are made for each room it will print out the total usable area of a house to the output screen. Analysis This program will calculate the usable area in square feet of a house with a maximum of four rooms that are rectangular
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control. The classroom observed had all the necessary areas of a preschool room. The room had a house keeping/dramatic play area as well as a block area‚ writing‚ science‚ computer‚ music‚ reading‚ circle‚ manipulative and art areas. All of the areas were appropriately spaced out and comfortable for the children. The house keeping/dramatic play area was the first visible area one would see when entering the class. This area includes a play stove and refrigerator with plastic food.
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1. What is the relationship between rate of ion exchange and surface-to-volume ratio? - As the surface-to-volume (SA:V) decreases the rate of ion exchange will also decrease. 2. Why is it important for a cell to have a large surface-to-volume ratio? - If the SA:V is too small then then rate of ion exchange will decrease‚ if this rate of ion exchange is too low then the cell will no longer be viable. 3. Which is more efficient at exchanging materials‚ a small or a large cell? EXPLAIN
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[Lihat sebelah SULIT SULIT 3 1449/1 SHAPES A D SPACE 1 2 3 4 5 6 7 8 Area of trapezium = 1 × sum of parallel sides × height 2 Circumference of circle = πd = 2πr 2 Area of circle = πr Curved surface area of cylinder = 2πrh 2 Surface area of sphere = 4πr Volume of right prism = cross sectional area × length Volume of cylinder = πr 2h Volume of cone = 1 2 πr h 3 4 3 πr 3 1 × base area ×
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