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Math 63 Section 1.1: Operations with Real Numbers Section 1.1 1. 16 4.8 x 105 4.3 x 10-4 7.5 x 103 4.2 x 10-8 6.2 x 107 16. 7" 16 2. A = 12.672 in C = 7.134 in B = 3.934 in D = 2.67 in 7" 8 3" H= 16 5" R = 20 8 3. S = 1 4. maximum 17.351 in 18.27 cm 21 26 32 in 320.06 cm 17. Width between centers = 58 1" 8 1" 31 2 3" 4 5. Width = 76 Height = 6. a) 7" 16 18. 3" 16 15" B = 5 16 3" C = 11 8 7. A = 3 1" 16 5" 104 8 8. Total Rise = 68 Total Run = 9. A = 109 3" 16 7" , 8 B = 122 this design will produce 84 parts. 10. A = 102.506 in B = 75.39 in, this design will produce 96 parts. 23 11. Area = 89 32 square inches Perimeter = 39 12. 13. 14. 23" Width = 2 32 1" Height = 1 8 11" A = 2 16 1" B=3 8, 3" 32 4 1" 4 15. a) 6,320,000,000 V b) .0000000098 A c) .00000047 F d) 14,200,000 W 480 kHz 430 µV 7.5 kW 42 nF 62MH minimum 17.283 in 17.97 cm 19 26 32 in 319.94 cm Difference 9 - 32 in - .05 mm + .041 in 7" b) Yes there is 8 to spare, although this is not much when divided among 9 spaces. 480 x 103 430 x 10-6 7.5x 103 42 x 10-9 62 x 106 19. 27 Rows, 41 Columns, this design will produce 1107 parts. 20. 32 Rows, 35 Columns, this design will produce 1120 parts. 21. Design Length Tolerance 3.7 in ± 36.62 in ± 80.5 in ± 100.29 in ± 2 100 2 100 2 100 2 100 Fraction 70 in 3 in 36 in 80 in 100 100 Max 72 in 3 in 36 in 80 in 100 62 100 50 100 29 100 100 Min 3 in 36 in 80 in 100 64 100 52 100 31 100 3 4 in 1 1 in 4 5 8 in 8 8 5 16 in 1 15 in 8 1 20 in 4 Tolerance 1 7 8 8 ± in 1 ± in 8 1 ± in 8 ± Maximum Dimension 1 16 in 1 ± in 8 1 ± in 8 in 3 1 in 8 3 8 in 4 3 8 in 8 1 15 in 4 3 20 in 8 Minimum Dimension 5 8 100 in 60 100 48 100 in in 27 100 22. Design Dimension 68 in in 1 1 in 8 1 8 in 2 1 8 in 4 15 in 1 20 in 8 in Math 63 Section 1.1: Operations with Real Numbers 5 8 1 4 1. A carpenter needs a board that is 79 " long that she will cut from a board is 96 " long. If the saw blade has a 3 " kerf blade (kerf is the thickness of wood that the blade removes with a 16 cut), calculate the length of the left over piece. Approach: Step 1. Add the length of the board needed (79-5/8”) to the kerf (3/16”). Step 2. Subtract the sum in step one from the original length of the board. Step 1 – Total of kerf plus length of board needed (T) Step 2 – Length of Board Leftover (L) 13 3 16 16 13 T 79 in 16 1 13 L 96 79 4 16 7 L 16 in 16 T 79 2. Calculate the missing dimensions A, B, C, and D. Measurements are in inches. Find A Approach: Add the two dimensions (3.247 in and 9.425 in) which make up the length of A. A 3.247 9.425 A 12.672 in Find B Approach: Step 1. Add the two dimensions on the same side as B (4.868 in and 3.870 in). Step 2. Subtract the sum in step 1 from the length found for A (12.672 in). B 12.672 4.868 3.870 B 12.672 8.738 B 3.934 in Find C Approach: C is the same length as its opposite side; total the three dimensions that make up the length of the opposite side (1.844 in, 2.721 in, 2.569 in). C 1.844 2.721 2.569 C 7.134 in Find D Approach: Subtract 4.464 in from the length found for C (7.134 in). D 7.134 4.464 D 2.67 in Math 63 Section 1.1: Operations with Real Numbers 3. Calculate the length of the nonthreaded portion of the bolt (S). Calculate the size of the overhang (H). Measurements are in inches. Find S Approach: Subtract the length of the threaded portion of the bolt (3-1/4 in) from the length of the length of the bolt (5-1/8 in). 1 1 S 5 3 8 4 7 S 1 in 8 Find H Approach: 7 1 Step 1. Subtract the diameter of the threaded portion of the bolt (1/2 in) from H 2 8 2 the diameter of the head of the bolt (7/8 in). Step 2. To get the size of the overhang, divide the answer from step 1 by two. 3 H 2 8 3 H in 16 4. If the drawing of the part below is scaled up by a factor of five, find the new dimensions. Measurements are in inches. Find the new Radius (R) 1 8 Approach: Multiply the 4 in radius by 5. 1 R 4 5 8 5 R 20 in 8 Find the new distance between the centers of the circles (D) 3 4 Approach: Multiply the distance 11 in by 5. 3 D 11 5 4 3 D 58 in 4 Math 63 Section 1.1: Operations with Real Numbers 5. If the drawing of the link below is scaled up by a factor of seven, find the new dimensions for the overall width and height. Measurements are in inches. Note: An R in drafting denotes the radius of a circle. Find the new overall width (W) Approach: Find the new overall height (H) Approach: 1. Multiply the radius 1. Multiply the radius 1 2 in by 2 and add the 4 distance between the centers of the circles). 2. Multiply the answer from step 1 by 7. 3 1 W 2 26 7 8 4 3 1 W 4 6 7 8 2 7 W 10 7 8 1 W 76 in 8 1 2 in by 2. 4 2. Multiply the answer from step 1 by 7. 1 H 2 27 4 1 H 4 7 2 1 H 31 in 2 Math 63 Section 1.1: Operations with Real Numbers 6. a) A fence is to have seven boards between two posts so that the space between each board is the same. Calculate the distance between each board. Answer as a fraction. b) If the carpenter decided the distance between boards was too large, would it be possible to add an extra board, decreasing the space between each board? Explain. Note: dashes are commonly placed between whole numbers and 7 8 fractions so 21 " is not mistaken for 217 " . 8 a) Find the distance between each board (D) Approach: 5 7 D 21 7 2 8 8 8 3 7 D 21 18 8 8 8 Step 3. Divide the answer from step 2 by 8 (because there are eight spaces D 3 1 8 between the boards and posts). 2 7 D in 16 5 Step 1. Multiply 7 boards times the width of each board 2 in . 8 7 Step 2. Subtract the answer from step 1 from the total length 21 in . 8 b) Can you add an extra board? Approach: 5 8 Step 1. Multiply 8 boards times the width of a single board 2 in and 7 compare the answer to the overall length of 21 in . 8 5 8 2 21 in 8 Step 2. If you can add the extra board, see how much space there would be leftover to allow for spacing between the boards and posts. There is enough room for 8 boards, but this only leaves That’s less than 7 in to be spread across 9 spaces. 8 1 in between each board. Although possible, this really isn’t very much room. 8 Math 63 Section 1.1: Operations with Real Numbers 7. Calculate the missing dimensions A, B, and C. Measurements are in inches. Find A Approach: 1. Total the dimensions of the side 5 1 3 opposite A 2 in, 2 in, 2 in . 8 2 4 2. Subtract the sum of the two known dimensions, on the same side as A, from the answer in step 1. 5 1 7 1 3 A 2 2 2 2 2 8 2 16 4 4 7 11 A 7 4 8 16 3 A3 in 16 Find B Approach: Multiply the radius 7 3 1 in by 2, then add 2 in . 16 4 3 7 B 2 1 2 4 16 1 7 B3 2 2 16 15 B5 in 16 Find C Approach: Add 2 times the radius 3 1 in plus all three dimensions on the 4 5 1 3 same side as C 2 in, 2 in, 2 in . 8 2 4 3 3 5 1 C 2 1 2 2 2 4 4 8 2 1 7 C 3 7 2 8 3 C 11 in 8 Math 63 Section 1.1: Operations with Real Numbers 9 16 5 8 8. A stair stringer is cut with a 7 " rise and a 11 " run. Code stipulates that all rises must be the same height and all runs must be the same length to avoid being a trip hazard. Calculate the total rise and total run for the stringer with nine steps. Find Total Rise Find Total Run 9 Approach: Multiply 9 times a rise of 7 in . 16 Approach: Multiply 9 times a run of 11 in . Total Rise = 9 7 = 68 9 16 1 in 16 5 8 Total Run 9 11 =104 5 8 5 in 8 Math 63 Section 1.1: Operations with Real Numbers 9. A manufacturer wants to layout the parts on a sheet of steel in 14 rows and 6 columns with a ½” space between each row and column. Calculate the width (A) and length (B) of the sheet that will be needed to produce the parts. How many parts will this design produce? Measurements are in inches. Find the width (A) Approach: 1. Find the width of 6 parts by multiplying the 5 15 in + 2 4 in 32 32 width of one part 9 times 6. 2. Find the length of all the spaces by multiplying 5 times 1 in . 2 3. Find the total width (A) by adding the answers from steps 1 and 2. 4. 5 1 15 A 9 2 4 6 5 32 2 32 10 1 15 A 9 8 6 2 32 2 32 1 25 A 17 6 2 2 32 11 1 A 106 2 16 2 3 A 109 in 16 How many parts will this design produce (N)? This design will produce a maximum of 84 parts. Find the length (B) Approach: 5. Find the length of 14 parts by multiplying the length of one part 2 4 5 in 32 times 14. 6. Find the length of all the spaces by 1 2 multiplying 13 times in . 7. Find the total length (B) by adding the answers from steps 1 and 2. 5 1 B 2 4 14 13 32 2 5 1 B 8 14 6 16 2 3 1 B 116 6 8 2 7 B 122 in 8 N 6 14 N 84 parts Math 63 Section 1.1: Operations with Real Numbers 10. A manufacturer wants to layout the parts on a sheet of plastic in 12 rows and 8 columns with a .15-inch space between each row and column. Calculate the width (A) and length (B) of the sheet of that will be needed to produce the parts. How many parts will this design produce? Measurements are inches. Find the width (A) Approach: 4. Fid the width of 8 parts by multiplying the width of one part (12.682 in) times 8. 5. Find the distance of all the spaces by multiplying 7 times 0.15 in. 6. Get the total width (A) by adding the answers from steps 1 and 2. Find the length (B) Approach: 1. Fid the length of 12 parts by multiplying the length of one part (6.145 in) times 12. 2. Find the distance of all the spaces by multiplying 11 times 0.15 in. 3. Get the total length (B) by adding the answers from steps 1 and 2. A 12.682 8 0.15 7 A 101.456 1.05 B 6.145 12 0.15 11 B 73.74 1.65 A 102.506 in B 75.39 in How many parts will this design produce (N)? This design will produce a maximum of 96 parts. N 12 8 N 96 parts Math 63 Section 1.1: Operations with Real Numbers 11. Calculate the area and perimeter of a rectangular sheet of glass that is 12-3/8” x 7-1/4”. Find the Perimeter (P) Approach: Use the formula P 2l 2w , where l = 12-3/8 in and w = 7-1/4 in l = length w = width P 2l 2 w 3 1 P 2 12 2 7 8 4 3 1 P 24 14 4 2 1 P 39 in 4 Find the Area (A) Approach: Use the formula A l w , where l = 12-3/8 in and w = 7-1/4 in l = length w = width Alw 3 1 A 12 7 8 4 23 A 89 sq in 32 Math 63 Section 1.1: Operations with Real Numbers 12. If the drawing of the link below is scaled down by a factor of four (divide by four), find the new dimensions for the overall width and height. Measurements are in inches. Find the new height (H) Approach: 3. Determine the current height. 4. Divide the answer from step 1 by 4. Find the new width (W) Approach: 1. Determine the current width. 2. Divide the answer from step 1 by 4. 1 H 22 4 4 1 H 4 4 2 1 H 1 in 8 1 3 W 22 6 4 4 8 3 1 W 4 6 4 8 2 7 W 10 4 8 23 W2 in 32 Math 63 Section 1.1: Operations with Real Numbers 13. A steel plate is to have five 7/8” diameter holes drilled so that the space between the edges of each hole and between the edge of each hole and the edge of the plate is the same. Calculate the distance from the edge of the plate to the center of the first hole (A), then the distance between the center of each hole (B). Answer as a fraction. Find the distance from the edge of the plate to center of the first hole (A) Approach: 1. Determine the total of the 5 diameters (each 7/8 in). 2. Subtract the value from step 1 from the total length (17-7/8 in). 3. Divide the answer from step 2 by 6 (6 is the number of spaces between the five holes (4) plus the two spaces from the outside holes to the edge). This will give the distance from the edge of the plate to the closest edge of the circle (also the distance between each hole). 4. Add half the diameter to the answer from step 3 in order to get the distance from the edge of the plate to the center of the first hole (A). 7 7 1 7 A 17 5 6 2 8 8 8 3 7 7 A 17 4 6 8 16 8 1 7 A 13 6 2 16 1 7 A2 4 16 11 A2 in 16 Find the distance between the centers of each hole (B) Approach: Use the answer from Step 3 in finding A and add two times the radius (7/8 in). 1 7 B2 4 8 1 B 3 in 8 Math 63 Section 1.1: Operations with Real Numbers 14. The face frame of a cabinet is made of vertical stiles and horizontal rails. Calculate the width of the rails in the design below so that all nine rails are the same length. Find the width of a single rail (W) Approach: 1. Find the total width of all of the stiles. 2. Subtract the result from step 1 from the total width (107 in). 3. Divide the result from step 2 by 3 to get the width of a single rail. 3 3 5 5 W 107 2 1 1 2 3 4 4 8 8 3 W 107 8 3 4 1 W 98 3 4 3 W 32 in 4 Math 63 Section 1.1: Operations with Real Numbers 15. In electronics large or small quantities are written using metric prefixes. For example: 3.2 μA = 3.2 x 10-6 A = 3.2 x .000001 A = .0000032 A 8.3 kΩ = 8.3 x 103 Ω = 8.3 x 1000 Ω = 8300 Ω otice that the power of 10 moves the decimal left or right according to the sign of the number. Common Electronics Units Unit Symbol W Watt Ω Ohm V Volt A Amp F Farad H Henry Hz Hertz Expand the following without prefixes: a) 6.32 GV b) 9.8 nA c) .47 μF d) 14.2 MW Metric Prefix Tera Giga Mega kilo milli micro nano pico Table of Common Metric Prefixes Symbol Power of 10 Factor T 1012 1,000,000,000,000 G 109 1,000,000,000 M 106 1,000,000 3 k 10 1,000 m 10-3 .001 -6 μ 10 .000001 n 10-9 .000000001 -12 p 10 .000000000001 Approach: Change the metric prefix to its factor equivalent, then multiply by the value given. Note: You can go directly from the exponential notation to the final value by moving the decimal in accordance with the value of the exponent. If the exponent is positive, move the decimal right. If the exponent is negative, move the decimal left. This is a much faster approach. 6.32 GV 6.32 109 V a) 6.32 1, 000, 000, 000 V .47 μF .47 106 c) = .47 0.000001 = 6,320, 000, 000 V = 0.00000047 F 14.2 MW 14.2 106 9.8 nA 9.8 109 b) = 9.8 0.000000001 = 0.0000000098 A d) = 14.2 1, 000, 000 = 14, 200, 000 W Math 63 Section 1.1: Operations with Real Numbers 16. Scientific Notation is a number between 1 and 10 times a power of ten. Engineering Notation is a modification of Scientific Notation where the number is between 1 and 1000 and the power of 10 is always a multiple of three, making it easy to write the number using a metric prefix. Consider the two examples in the table and fill in the rest using the table of common metric prefixes. Metric Prefix Tera Giga Mega kilo milli micro nano pico Standard Form 12,300 Ω .000072 A 480,000 Hz .00043 V 7,500 W .000000042 F 62,000,000 H Table of Common Metric Prefixes Symbol Power of 10 Factor 12 T 10 1,000,000,000,000 G 109 1,000,000,000 6 M 10 1,000,000 k 103 1,000 -3 m 10 .001 μ 10-6 .000001 n 10-9 .000000001 p 10-12 .000000000001 Scientific Notation 1.23 x 104 7.2 x 10-5 4.8 x 105 4.3 x 10-4 7.5 x 103 4.2 x 10-8 6.2 x 107 Engineering Notation 12.3 x 103 72 x 10-6 480 x 103 430 x 10-6 7.5 x 103 42 x 10-9 62 x 106 Metric Prefix 12.3 kΩ 72 μA 480 kHz 430 μV 7.5 kW 42 nF 62 MH Math 63 Section 1.1: Operations with Real Numbers 17. Consider a rod being cut to a length. The length can never be perfect and so must be cut for accuracy based on a desired tolerance. Tolerance is the amount that a measurement can vary above or below a design size. Approach: For maximum length, add the tolerance to the design length. For minimum length, subtract the tolerance from the design length. Note: for the last problem in the table, first change meters to centimeters, then apply the tolerance as before. (1 meter = 100 centimeters, therefore, 3.2 meters = 320 centimeters) Design Length Tolerance Maximum Length Minimum Length 17.317 in ± .034 in 17.351 in 17.283 in 18.12 cm ± .15 cm 18.27 cm 17.97 cm 5 ± 26 8 in 3.2 meters 1 32 in ± .06 cm 26 21 in 32 320.06 cm 26 19 in 32 319.94 cm 18. Calculate the difference between the design size and the actual size in the chart. Actual sizes that are less than the design size should be denoted with negative numbers. Approach: Subtract the actual size from the design size. 14 3 29 13 16 32 Design Size 3 14 16 in Actual Size 13 29 32 in Difference 9 in 32 18.43 18.38 18.43 mm 18. 38 mm -0.05 mm 17.972 in 18.013 in 0.041 in 17.972 18.013 Math 63 Section 1.1: Operations with Real Numbers 19. Challenge problem: A manufacturer wants to layout parts on a sheet of steel that is 120” x 68”. If the part is 2.34” wide and 1.87” tall, calculate the number of rows and columns that will fit if there is a .6” space between each part. How many parts will this design produce? Note: the drawing is not to scale. Find the number of parts this design will produce (N) Approach: 1. Find the number of parts (A) that can fit horizontally (120 inches) if each part is 2.34 in long. 2. Find the number of parts (B) that can fit vertically (68inches) if each part is 1.87 in wide. 3. Multiply the answers from steps 1 and 2 to find the total number of parts (N). Note: There is no space between the part and the edge of the sheet. There are only gaps between the parts. Therefore, there will be one less gap than there are parts (e.g. 15 parts across means 14 gaps). Step 1. Find the number of parts horizontally A = number of parts Step 2. Find the number of parts vertically B = number of parts Each part is 2.34 plus a gap of 0.6, but we need to subtract the gap at the end. 120 (2.34 0.6) A 0.6 120 2.94 A 0.6 120.6 2.94 A Each part is 1.87 plus a gap of 0.6, but we need to subtract the gap at the end. 68 (1.87 0.6) B 0.6 68 2.47 B 0.6 68.6 2.47 B 120.6 A 2.94 A 41.02 parts 68.6 B 2.47 B 27.77 parts Therefore, the maximum number of parts horizontally is 41. Therefore, the maximum number of parts vertically is 27. Step 3. The maximum number of parts this design will produce is 1107. N 41 27 N 1107 parts Math 63 Section 1.1: Operations with Real Numbers 20. Challenge Problem: A manufacturer wants to layout parts on a sheet of steel that is 108” x 60”. If the part is 2 5" wide and 16 1 1" 8 tall, calculate the number of rows and columns that will fit if there is a 3" 4 space between each part. How many parts will this design produce? Note: the drawing is not to scale. Find the number of parts this design will produce (N) You can use fractions or decimals to calculate. Approach: 1. First find the number of parts (A) that can fit horizontally (108 in) if each part is 2-5/16 in long. 2. Find the number of parts (B) that can fit vertically (60 in) if each part is 1-1/8 in wide. 3. Multiply the answers from steps 1 and 2 to find the total number of parts (N). Note: There is no space between the part and the edge of the sheet. There are only gaps between the parts. Therefore, there will be one less gap than there are parts (e.g. 15 parts across means 14 gaps). Step 1. Find the number of parts horizontally A = number of parts, A – 1 = number of gaps Step 2. Find the number of parts vertically B = number of parts, B – 1 = number of gaps 3 5 A 1 2 A 4 16 3 3 5 108 A 2 A 4 4 16 1 3 108 3 A 16 4 3 1 108 3 A 4 16 3 1 108 3 A 4 16 25 A 35 parts 49 60 108 3 1 B 1 1 B 4 8 60 0.75 B 1 1.125B 60 0.75B 0.75 1.125B 60 1.875B 0.75 60.75 1.875B B 32.4 parts Therefore, the maximum number of parts vertically is 32. Therefore, the maximum number of parts horizontally is 35. Step 3: The maximum number of parts this design will produce is 1120. N 35 32 N 1120 parts Math 63 Section 1.1: Operations with Real Numbers Math 63 Section 1.1: Operations with Real Numbers 21. Consider the diagram. Convert the given 1 dimensions to fractions to the nearest 100 inch (do 2 not simplify). Using a tolerance of ± 100 inch, determine the maximum and minimum allowable dimensions. Approach: 1. To rewrite the design length as a fraction, take the decimal out two places (if necessary), then rewrite as a mixed number. 2 to the mixed number. Do not simplify. 100 2 3. To get the minimum length, subtract from the mixed number. Do not simplify. 100 2. To get the maximum length, add Design Length Tolerance 3.7 in 2 in 100 3 16.62 in 2 in 100 80.5 in 100.29 in Fraction Maximum Length 70 100 3 16 62 100 2 in 100 80 2 in 100 100 Minimum Length 72 100 3 68 100 16 64 100 16 60 100 50 100 80 52 100 80 48 100 29 100 10 31 100 10 27 100 Math 63 Section 1.1: Operations with Real Numbers 22. Consider the diagram. Determine the maximum and minimum allowable dimensions for the given tolerances in the table below. Answers should be simplified. Note: the symbol Ø indicates the diameter of the circle. Side View End View Approach: 1. To get the maximum length, add the tolerance to the design dimension. 2. To get the minimum length, subtract the tolerance to the design dimension Note: All answers should be fully simplified. Design Dimension Tolerance 3 4 1 1 4 5 8 8 5 8 16 1 15 8 1 20 4 1 in 8 1 in 8 1 in 8 1 in 16 1 in 8 1 in 8 Maximum Minimum Dimension Dimension 7 8 3 1 8 3 8 4 3 8 8 1 15 4 3 20 8 5 8 1 1 8 1 8 2 1 8 4 15 20 1 8