NOTE: Use the Trapezoidal rule for all irregular shaped questions. Assume measurements are taken in order (first to last) unless otherwise stated.
1. An artificial green needs to be covered in turf, and the gardener in an effort to calculate exactly how much is needed, measures the lengths across the green at 1 metre intervals. The gardener measures lengths of 1, 3, 9, 4, 5, 6 and 8 metres respectively. What area needs to be covered?
2. A parade float is in the shape of an island. The designer decides to cover the island in green cloth to make it look more realistic, and measures the width of the float every 50 centimetres, and records the first length as 1.5 metres, and the others as 2.5, 3, 4, 3, 2.5, 3 and 1 metre respectively. How much cloth will be placed on the float?
3. A section of a roller coaster's tracks needs to have a cover installed to put up a sign. The track on this section has no curves, but does go up and down. The engineers measure each of the nine supports, which are spaced 3 metres apart, and record the following lenghts in order: 4, 5, 6, 3, 2, 3, 4, 5, 5 (in metres). What area will the sign cover?
4. An artist designed abstract pool needs to be covered in case of rain, and other bad weather. The pool is measured in intervals of 30 centimetres and has lengths of 0m, 2m, 1m, 2m, 3m, 4m, 3m and 1 m, where m is metres. What is the surface area of the cover?
5. A satellite photo of a parcel of land shows that the distances from one side of the resort, which is 500 metres long, to the other, is 0 metres, 100 metres, 320 metres, 200 metres, 150 metres, 200 metres, 400 metres, 350 metres, 250 metres, 200 metres and 10 metres. How large is the parcel of land, given that the distances are measured at even intervals?
6. A hole in the side of a blimp needs to be patched. How much material is needed if the gap, measured at 1 foot intervals, is 3 feet, 2 feet, 4 feet, 4 feet, 5 feet, 3 feet, 2 feet and 1 foot?
7. Trinidad Asphalt is loading asphalt to fill a pot hole that is 1 foot, 6 feet, 3 feet, 4 feet, and 2 feet wide at points in the pothole, which are 1 foot apart from each other. What surface area does the pothole have?
8. A man man decides to paint the town red, but because of the global economic recession, wants to make sure he is as efficient as possible in the purchasing of his paint. If the town is measured at 100 metre intervals and the following distances (in metres) are found respectively: 500, 900, 850, 450, 700, 800, 850, 900, 850, 900, 1000, 1250, 1100, 900, 800, 750. What surface area would he need to paint?
9. A tee-shirt printer needs to purcahse just the right amount of ink to print a splash on a tee shirt. The splash is measured at 1cm intervals and is 1cm high at one end, 2cm at the other, and the lengths in between are 2cm, 3cm, 4cm, 2cm, 3cm, 5cm, 3cm and 2cm.
10. A footprint is seen at a crime scene and investigators want to find out the surface area to link it to another suspicious footprint they have found. They measure the following lengths, at 1 inch intervals, respectively. and find 2.5 inches, 2.8 inches, 2.5 inches, 2.4 inches, 2.2 inches, 2.2 inches, 2.0 inches, 2.4 inches, 2.5 inches, 2.6 inches and 2.7 inches. What is the area of the footprint?
11. An oil barrel measures 4 feet in diameter and is 6 feet high, how much oil can it hold?
12. A grain silo is in the shape of a cylinder and is placed on a square base that fits it perfectly with no overlap of any kind. The base measures 6 metres along one side. If the silo is 20 metres tall, how much grain can it hold?
13. You think a can of Juice is rather small for it to hold the 2000 centimetres cubed that it says on the label. You measure the can and get the following dimensions: Height: 15 centimetres, Diameter: 5 centimetres, Circumference: 15.714 cm. How much juice does the can actually hold?
14. How much water can your dog's bowl hold if it has a radius of 6 inches and a height of the same?
15. A programmer needs to tell the computers how much grease to fill in a cylindrical tube of height 30 centimetres and diameter 6 centimetres. How much will he need to tell the computers to fill the tubes?
16. If a coin measures 2cm in diameter and .1 cm in thickness, and ten of these are placed in a tub of water that is full to the brim, how much cm³ will the coins cause to spill?
17. An absorbant rope is used to trace the edge of an oil spill. Measuring the rope at 1.5 metre intervals, the distances across the spill, in order, is found to be 1m, 3m, 10m, 20m, 15m, 10m, 15m, 20m, 10m, 5m, 10m, and 2m. How much oil has spilled?
18. A cupboard door needs to be lined with insulating material, and the door measures 30cm, 40cm, 45cm, 45cm, 45cm, 40cm, 35cm, 30cm and 20cm across, at points which are 12 cm apart. How much insulation is needed?
19. A patch of grass in a field is dead, determine how much grass died if you are told that the area was measured in gaps of 32 inches and the distances across were found to be 40 inches, 4 feet, 5 feet, 53 inches, 6 feet, 50 inches and 23 inches.
20. A hole needs to be patched. The following measurements are taken at 1.25 cm intervals: 1cm, 2.20cm, 2.10cm, 1.75cm, 1.44cm, 1.20cm, 1.0cm, 1cm, .55cm and 0cm. How much area needs to be patched?
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SOLUTION FOR QUESTION 1
ReplyDeleteextracting the info given
1st ordinate=1
last ordinate,n=8
remaining ordinates=3, 9, 4, 5, 6
width of strip,b=1
trapezoidal rule=width of strip(1/2 sum of first +last ordinates)+ sum of reminding ordinates
= 1(1/2[1+8])3+9+4+5+6
=1*4.5+3+9+4+5+6
=31.5m sq
solution for question 2
ReplyDeletefrom question info given:
1st ordinate=1.5
last ordinate= 1
remaining ordinates= 2.5,3,4,3,2.5,3
using the trapezoidal rule:
width of the strip[(1/2)(sum of the 1st and last ordinates)+(sum of the other ordinates)]
=(0.5)[(1/2)(1.5+1)+(2.5+3+4+3+2.5+3)]
=(0.5)*[1.25+18]
=(0.5)*(19.25)
=9.63m sq
Q.19
ReplyDeleteusing the Trapezoidal rule= width of strip [1/2 (sum of the first an the last ordinates)+the sum of the remaing ordinates]
32[1/2(40+23)+(48+60+53+72+50)]
=10432 inches sq
solution for Q3.
ReplyDeleteUsing thetrapezoidal rule:
[1/2(the sum of the first and last ordinates)+(the sum of the remaining ordinates)]
The first ordinate: 4
The second ordinate: 5
=[3(1/2)(4+5)+(5+6+3+2+3+4+5)]
=97.5m sq.
(1) trapezium rule:
ReplyDeleteb/2(1st+last ordinate)+(sum of remaining
ordinates)
therefore
1/2(1+8)+(3+9+4+5+6)
=1/2(9)+(27)
=1/2(36)
=18m^2
Question 4
ReplyDeleteUsing thetrapezoidal rule:
width of strip(.5(the sum of the first and last ordinates)+(the sum of the remaining ordinates)]
30(0+1)+(2+1+2+3+4+3)
30(.5(1)+(15)
30(8)
240 m ssq.
Using thetrapezoidal rule:
ReplyDeletewidth of strip(.5(the sum of the first and last ordinates)+(the sum of the remaining ordinates)]
1(.5(4)+(2+4+4+5+3+2)
1(2)+(20)
22 feet sq.
(2) width=50/100
ReplyDelete=.5m
using trapezium rule
1/2(1.5+1)+(2.5+3+4+3+2.5+3)
=1/2(2.5)+(18)
=1/2(20.5)
=10.25m^2
solution for question 14..
ReplyDeletesince the radius is 6 inches and the height is 15 centimetres
then:
circumference= 3.14x(6^2)
= 113.04
To find volume multiply area of circle by height
= 113.04x6
= 678.24
Trapezoidal rule
ReplyDeletewidth of each strip Area of the first trapezium + Area of the 2nd trapezuim
1 * 1/2 (1 +8) + 3 + 4 + 9+ 5+ 6
1/2 (9) + 27
31.5 m
Question 2
ReplyDeletew = 50 cm needs to be converted to metres so 50/100 = 0.5m
where w is the interval or width of each strips.
0.5 * 1/2 (1.5 + 1 ) + 2.5 + 3 + 4 + 3 + 2.5 + 3)
so 1/2 multpily by 1.5 + 1 = 2.5 = 1.25
0.5 (1.25) + 18
0.5 * 19.25
9.6 m
Question 11
ReplyDeleteVolume of cylinder-22/7*r^2*h
3.142(2)^2(6)
=75.41 ft^3
Question 14
ReplyDeleteVolume of cylinder-22/7*r^2*h
3.142(6)^2(6)
3.142(36)(6)
=678 inches^3
Using thetrapezoidal rule:
ReplyDeletewidth of strip(.5(the sum of the first and last ordinates)+(the sum of the remaining ordinates)]
1(.5(1+2)+(6+3+4)
1(.5(3)+(13))
1(.5(16))
8 ft ^2
Above Question was No.7
ReplyDeleteQuestion 4
Using thetrapezoidal rule:
width of strip(.5(the sum of the first and last ordinates)+(the sum of the remaining ordinates)]
30(.5(0+1)+(2+1+2+3+4+3)
30(.5(1)+(15)
30(8)
=240 m ^2
QUESTION 11 :An oil barrel measures 4 feet in diameter and is 6 feet high, how much oil can it hold?
ReplyDeleteArea of a circle = 22/7 * r ^2
Area of a side of cylinder = 2 * 22/7 * r * h
Raduis = diameter / 2
4 / 2 = 2
2 * 22/7 * 2*6
= 75.4
Someone please add the units I am not quite sure ; Help Help
Solution for Q6.
ReplyDeleteUsing the trapezoidal rule:
[The width of the strip(1/2)(the sum of the first and last ordinate)+(The sum of the remaining ordinates)].
Width of the strip is: 1 sq.ft.
The first ordinate is: 3 sq.ft.
The last ordinate is: 1 sq. ft.
The remaining ordinates is: [2,4,4,5,3,2].
subsituting into the formula:
= [1*(3+1)+(2+4+4+5+3+2)].
= [1*4(20).
= [4(20).
= 80 sq.ft
Qu 12 : A grain silo is in the shape of a cylinder and is placed on a square base that fits it perfectly with no overlap of any kind. The base measures 6 metres along one side. If the silo is 20 metres tall, how much grain can it hold?
ReplyDeleteArea of a square = s ^ 2
6^2 = 36
So if it is 20 m tall you say 20 * 36 = 720 m
I am not sure on my reasoning.
Correct if I am wrong
solution 14
ReplyDeletearea of circle=pie radius(sq)
area of circle= 3.14*6*6
area of circle= 113.04
so
volume is area multiple height
height=6
area=113.04
volume=113.04*6
volume=678.24
Solution for Q.20.
ReplyDeleteUsing the trapezidal rule:
[The width of the strip(1/2)(the sum of the first and last ordinates)+ (The sum of the remaining ordinates)].
Width: 1.25 cm.
The first ordinate is: 1cm.
The last ordinate is: 0cm.
The remaining ordinates is:[2.20,2.10,1.75,1.44,1.20,1,1,55].
subsituting into the formula:
[1.25(1+0)+(2.20 + 2.10 +1.75 + 1.44 + 1.20 + 1.0 + 1 + 55)].
= [1.25 + 65.69].
= 66.94 cm.
= 67 cm.
Question 20 : using the trapezodial rule which states that area of an irregular figure can be found by : A = width of each strip * 1/2 sum of the first + last ordinate + sum of the remaining ordinates.
ReplyDelete1.25 * 1/2 ( 1 + 0) + 2.20 + 2.10 + 1.75 + 1.44 + 1.20 + 1.0 + 1 + 55
question 2
ReplyDeletewidth=50cm
which is = to 0.5m
0.5*1/2(1.5+1)+2.5+3+4+3+2.5+3
=0.5*(1.25+18)
=0.5*(19.25)
=9.6m^2
Solution for Q.18.
ReplyDeleteusing the trapezoidal rule:
[width of strip(1/2)(The sum of the first and lass ordinates) + (The sum of the remaining ordinates).
Width = 12 cm.
First ordinate = 30 cm.
Last ordinate = 20 cm.
subsituting into the formula.
[12(1/2)(30 + 20) + (40 + 45 + 45 + 45 + 40 + 35 + 30)
=(300 + 280)
= 580 cm.
Solution for Q.17.
ReplyDeleteUsing the trapezoidal rule:
[width of strip(1/2)(the sum of the first and last ordinate)+(the sum of the remaining ordinates).
width = 1.5 m
first ordinate = 1 m
last ordinate = 2 m
subsituting into the rule:
[1.5(1/)(1 + 2)+(1+3+10+20+15+10+15+20+10+5+10)]
= (2.25 + 119)
= 121.25 m.
question 20.
ReplyDeletewidth intreval = 1.25cm
lengths of strips = 1cm, 2.20cm, 2.10cm, 1.75cm, 1.44cm, 1.20cm, 1.0cm, 1cm, .55cm and 0cm
using the trpezoidal rule
A = width of strip[1/2(sum of the first and last ordinates)+ (sum of the remaining ordinates)
A =1.25[1/2(1+0)+ (2.2+2.1+1.75+1.44+1.2+1+1+.55)
= 1.25 [0.5+ 11.04]
= 14.425cm^2
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteusing trapezoidial rule
ReplyDeleteA = 32[1/2(40+23)+(48+60+53+72+50)
= 32[31.5+ 283]
= 32(314.5)
area of grass died = 10064 square inches
Solution 1
ReplyDeletetrapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (1)x [(1/2 x(1+8)]+(3+9+4+5+6)
= (1)x (1/2 x 9)+(27)
= 1 x 4.5 + 27
= 31.5m2
Solution 2
ReplyDeletetrapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (.5)x[1/2(1.5 + 1)]+ (2.5+3+4+3+2.5+3)
= (.5)x[1/2(2.5)] + (18)
= (.5) x (1.25 + 18)
= (.5) x 19.25
= 9.63m2
Solution 3
ReplyDeletetrapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = 3 x [1/2 (4 + 5)] + (5+6+3+2+3+4+5)
= 3 x [1/2 (9)] + (28)
= 3 x (4.5+28)
= 3 x (32.5)
= 97.5m2
SOLUTION TO NO.8
ReplyDeleteinfo given
width,b=100m
first ordinate=500m
last ordinate,n=750m
remaining ordinates (in m)=900, 850, 450, 700, 800, 850, 900, 850, 900, 1000, 1250, 1100, 900, 800
Trapezoidal rule=width of strip(1/2 sum of
first +last ordinates)+ sum of reminding ordinates
=100[(1/2* 500+750)+900+ 850+ 450+ 700+ 800+850+900+ 850+ 900+ 1000+ 1250+ 1100+900+ 800
=100[(1/2*1250)+900+ 850+ 450+ 700+ 800+850+900+ 850+ 900+ 1000+ 1250+ 1100+900+ 800
=100[625+900+ 850+ 450+ 700+ 800+850+900+ 850+ 900+ 1000+ 1250+ 1100+900+ 800]
=100*12875
=1287500m sq
if the answer is off please correct it for me okay.
Solution 4
ReplyDeletetrapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (.3)x[1/2(0+1)] + (2+1+2+3+4+3)
= (.3)x(.5) + (15)
= (.3)x ( 15.5)
= 4.65m2
#11
ReplyDeleteArea of circle=22/7*r^2
=22/7*2^2
=12.6 feet^2
volume of cylinder=area of triangle*h
=12.6*6
=75.6feet^3
solution 8
ReplyDeletetrapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (100)x[1/2 (500 + 750)]+(900+850+450+700+800+850+900+850+900+1000+1250+ 1100+900+800)
= (100)x [1/2(1250)] + (12250)
= (100)x (625 + 12250)
= 100 x 12875
= 1287500m2
i got the same answer THE DAWG.....
#12
ReplyDeleteArea of circle=22/7*r^2
=22/7*3^2
=28.3m^2
Volume of cylinder=area of circle*height
=28.3*20
=556m^3
question 18
ReplyDeleteusing trapezoidial rule
A = 12[1/2(30+20)+(40+45+45+45+40+35+30)
= 12[ 25+280]
area of insulating material that is needed to line the cupboard door = 3660 squared cm
#13
ReplyDeleteArea of circle=22/7*r^2
=22/7*2.5^2
=19.6cm^2
Volume of cylinder=area of circle*height
=19.6*15
=294cm^3
Solution 10
ReplyDeletetrapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (1") x [1/2 (2.5" + 2.7")] + (2.8"+2.5"+2.4"+2.2"+2.2"+2.0"+2.4"+2.5"+2.6")
= (1") x [1/2 (5.2")] + (21.6")
= (1")x (2.6" + 21.6")
= 1" x 24.2"
area of foot print = 24.2"sq
SOLUTION 10
ReplyDeleteinfo given
width,b=1 inch
first ordinate=2.5inches
last ordinate,n=2.7inches
remaining ordinates (in inches)=2.8 inches, 2.5 inches, 2.4 inches, 2.2 inches, 2.2 inches, 2.0 inches, 2.4 inches, 2.5 inches, 2.6 inches
Trapezoidal rule=width of strip(1/2 sum of
first +last ordinates)+ sum of reminding ordinates
=1[(1/2*2.5+2.7)+2.8 + 2.5 + 2.4 + 2.2 + 2.2 , 2.0 + 2.4 + 2.5 + 2.6]
=1[(1/2*5.2)+2.8 + 2.5 + 2.4 + 2.2 + 2.2 , 2.0 + 2.4 + 2.5 + 2.6]
=1[2.6+2.8 + 2.5 + 2.4 + 2.2 + 2.2 , 2.0 + 2.4 + 2.5 + 2.6]
=1*24.2
=24.2"sq
Solution 17
ReplyDeletetrapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (1.5)x [1/2(1+2)] + (3m+10m+20m+15m+10m+15m+20m+10m+5m+10m)
= (1.5) x [1/2(3)] + (118)
= (1.5) x (1.5 + 118)
= 1.5 x 119.5
= 179.25m2
This comment has been removed by the author.
ReplyDeleteSOLUTION 15
ReplyDeleteinfo given
height=30 cm
diameter=6 cm
Volume of cylinder=cross section area * h
where
cross section area= pie*r^2
h=lenght of solid
=r= diameter/2=6/2=3cm
we can now calculate volume
=pie*3^2*30
=848.2cm^3
QUESTION 11
ReplyDeletediameter= 4 feet
height =6 feet
Volume of cylinder=cross section area * h
where
cross section area= pie*r^2
h=lenght of solid
r=d/2=4/2=2 ft
we can now calculate volume
=pie*2^2*6
=75.3 ft cube
question 17
ReplyDeleteusing simpsons rule
A = w/3[(sum of the first and last ordinates)+(2*sum of the remaining odd ordinates)+(4 times the remaining even ordinates)]
area = 1.5/3[3 + 116 + 240]
= 1.5/3(359)
= 179.5 m^2
question 16
ReplyDeletethe volume of one coin is = (pie*1^2)0.1
= (3.14)0.1
= (0.314)cm^3
volume of ten coins = 10 *0.314
= 3.14cm^3
the coins are solid so they will displace their own volume in water
therefore thevolume of water spilled = 3.14cm^3
question 15
ReplyDeletevolume of a cylinder =(pie * radius^2)h
=(pie * 3^2)30
= 28.27 * 30
= 848.1 cm^3
question 14
ReplyDeletevolume of a cylinder = (pie * radius^2)h
= (pie*6^2)6
= (113.1 * 6)
= 678.6 cubic inches
question 13
ReplyDeletevolume = (pie * radius^2)h
= (pie*2.5^2)15
= 19.63*15
actual volume of juice = 294.45 cm^3
question 12
ReplyDeletesince the silo is restin on top of the square.
we are only required to find the volume of the tower. since the tower touches the square from side too side the diameter of the tower is 6m.
therefore the volume of the tower =(pie *
radius^2)h
= (pie * 3^2)20
= 565.4 m^3
An oil barrel measures 4 feet in diameter and is 6 feet high, how much oil can it hold?
ReplyDeletevolume of oil barrel can hold = (pie * radius^2)h
= (pie * 2^2)6
= 12.57 *6
= 75.42 cubic feet
SOLUTION 18
ReplyDeletenfo given
width,b=12cm
first ordinate=30cm
last ordinate,n=20cm
remaining ordinates=40cm, 45cm, 45cm, 45cm, 40cm, 35cm, 30cm
Trapezoidal rule=width of strip(1/2 sum of
first +last ordinates)+ sum of reminding ordinates
=12[(1/2*30+20)+40+45+45+45+40+35+30
=12[25+40+45+45+45+40+35+30]
=12*305
=3660cm sq
This comment has been removed by the author.
ReplyDeletequestion 10
ReplyDeleteusing simpsons rule
A = w/3[(sum of the first and last ordinates)+(2*sum of the remaining odd ordinates)+(4 times the remaining even ordinates)]
A = 1/3[(2.5 + 2.7)+ (2*(2.8+2.4+2.2+)+(4(2.5+2.2+2.6)
= 1/3[5.2 + 14.8+ 29.2]
=1/3(49.2)
=16.4 square inches
Solution:1
ReplyDelete1st=1
last=8
remaining ordinates=3, 9, 4, 5, 6
width of strip=1
trapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (1)x [(1/2 x(1+8)]+(3+9+4+5+6)
= (1)x (1/2 x 9)+(27)
= 1 x 4.5 + 27
= 31.5m^2
Solution:2
ReplyDelete1st ordinate=1.5
Last ordinate=1
Remaining ordinates=2.5,3,4,3,2.5,3
W= 50cm -> 0.5m
trapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (0.5)x [(1/2 x(1.5+1)+(2.5+3+4+3+2.5+3)]
=(0.5)*[1.25+18]
=(0.5)*[19.25]
=9.63m^2
Solution:3
ReplyDelete1st ordinate=4
Last ordinate=5
Remaining ordinates=5+6+3+2+3+4+5
trapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (3)x [(1/2 x(4+5)+(5+6+3+2+3+4+5)]
= 3 x [(1/2x9) + (28)]
= 3 x [4.5+28]
= 3 x [32.5]
= 97.5m^2
Solution:4
ReplyDelete1st ordinate=0
Last ordinate=1
Remaining ordinates=2+1+2+3+4+3
W= 30cm -> 0.3m
trapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (0.3)x [(1/2 x(0+1)+(2+1+2+3+4+3)]
= (.3)x[(.5) + (15)]
= (.3)x [ 15.5]
= 4.65m^2
Solution:5
ReplyDelete1st ordinate=0
Last ordinate=10
Remaining ordinates=100,320,200,150,200,400,350,250,200,
W= 500m
trapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (500)x [(1/2 x(0+10)+(100,320,200,150,200,400,350,250,200)]
= (500)x[(5) + (2170)]
= (500)x [ 2175]
= 1087500m^2->1.08x10^6 m^2
Solution:6
ReplyDelete1st ordinate=3
Last ordinate=1
Remaining ordinates=2,4,4,5,3,2
W= 1 foot
trapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (1)x [(1/2 x(3+1)+(2,4,4,5,3,2)]
= (1)x[(2) + (20)]
= (1)x [22]
= 22 sq.ft
Solution:7
ReplyDelete1st ordinate=1
Last ordinate=2
Remaining ordinates=6+3+4
W= 1 foot
trapizoidal rule = (width of strip)(1/2 sum of 1st ordinates + last ordinates)+(sum of remainding ordinates)
Area = (1)x [(1/2 x(1+2)+(6+3+4)]
= (1)x[(1.5) + (13)]
= (1)x [14.5]
= 14.2 sq.ft
when de math project due?
ReplyDelete31st march
ReplyDeleteIs that a math 110D project? If so what is it.
ReplyDelete