Filimon's Volume Post

Pg 263 #5 v=h x pi x r2

r=4.1 11=h pi=3.14

r2 r x r = 16.81

Volume of this cylinder is 51.35.

#4 Pg 265

r2=10

h=23

pi=3.14

v=722.2

Roemer's Volume Scribe Post

7.3



















Height = 10cm
Diameter = 20.3cm

r= d/2 r = 20.3 / 2
r = 10.5cm

v = π x r x r x h
v = 3.14 x 10.15 x 10.15
v = 323.49cm2
v = 323.49 x 10
v = 3234.9cm3

7.4
















Height = 10m
Diameter = 0.8 (inside) and 1 (outside)

Inside -
r = d/2
r = 0.8/2
r = 0.4

v = π x r x r x h
v = 3.14 x 0.4 x 0.4 x h
v = 0.5 x 10
v = 5m

Outside -
r = d/2
r = 1/2
r = 0.5

v = π x r x r x h
v = 3.14 x 0.5 x 0.5 x h
v = 0.78 x 10
v = 7.8m

I tried to look for the volume of the concrete to build the Culvert. I basically just took the volume from the inside, and subtracted it from the volume in the outside.

7.8 - 5m = 2.8m3

Square Units

What you have to have on your green fold-able .
On one side .
Rectangular Prism : one cube with a net and a rectangle with a net .
Triangular Prism : 3 different triangles with one net for each one .
Opposite side .
Rectangular Prism

• all rectangular prisms have 6 faces
  
• opposite sides of each face are equal
  
• There are 3 views , TOP , FRONT and SIDE

Total Surface Area





















+ Also , make sure to add today's homework to the rest of the space on that back side .
Today's Homework

Paulo's Surface Area of a Prism

Homewor
pg 54, question 2

2. Calculate the surface area of each rectangular prism to the nearest tenth of a centimetre squared.








Front:
5x11.5=57.5cm2
5x11.5=57.5cm2

Top:
11.5x3.2=36.8cm2
11.5x3.2=36.8cm2

Side:
5x3.2=16cm2
5x3.2=16cm2
---------------
TSA 220.6cm2

b)
Front:
12x10.4=124.8cm2
12x10.4=124.8cm2

Top:
10.4x4.5=46.8cm2
10.4x4.5=46.8cm2

Side:
5x3.2=16cm2
5x3.2=16cm2
----------------
TSA 375.2 cm2

pg 54, question 3

3

a)
triangle+ 7x8/2= 28cm2
triangle+ 7x8/2=28cm2

a=2.5x8=20cm2
a=2.5x7=17.5cm2
a=2.5x=10.6=26.5cm2
------------------------
TSA 120cm2

pg 53, question 4




3.14x4.1x4.1)x11

Math Test Questions 1 and 3

Answers for questions 1 and 3 on the test

1

3

Here is a website for practice

http://www.thatquiz.org/tq-4/?-j2080-la

Cylinder Volume and Volume Problems

Jumbo

V= π.r.r.h

V= 3.14.10.10.40

V= 12 560cm³

Popcorn Lover's

V= π.r.r.h

V= 3.14.15.15.20

V= 14 130cm³

I think that Martha should get the Popcorn Lover's one because it has more volume to get more popcorn.

Inside:

V= π.r.r.h

V= 3.14 .0.4 .0.4 .10

V= 5.024cm³

Outside:

V= π.r.r.h

V= 3.14. 0.5 . 0.5 . 10

V= 7.85cm³

7.85 - 5.024=2.826m³

Surface Area- cylinders.

Cylinders:
The bases of the cylinder are circles.

Circumference:
-The perimeter of a circle .


Diameter:
- Halfs the circle.

Radius:
- Half of the diatmeter


*Formulas :
Finding the circumference:
πd= c
Finding the diamer :
2r=d or c/π= d
Finding the radius :
d/2= r

*Area of circle:
π x r 2
(3.14 x radius squared)


* Homework questions :)





A= πr²
A = 314 x 3.2 <sup>2
A= 3.14 x 10.24
A= 232.15 cm 2

A=</sup>π <sup>r2
A= 3.14 x 62
A= 3.14 x 36
A=113.04cm 2

d/2= r
25/2 = r
25/2= 12.5
A=</sup> π<sup>r2
A=3.14 x 62
A= 3.14 x 36
A= 490.526 cm 2

A=</sup>π<sup>r2
A= 3.14 x 16.5 2
A- 3.14 x 272.25
A= 854.865 cm 2

c/ =d
18.84/ 3.14=d
18.84/ 3.14= 6
d/2= r
6/2= r
6/2= 3
A=</sup> π<sup>r2
A= 3.14x 32
A= 3.14 x 9
A= 28.26 cm2

c/ = d
31.4/ 3.14= d
31.4/3.14= 10
d/2 = r
10/2= r
10/2 = 5
A=</sup> π<sup>r2
A= 3.14 x 52
A= 3.14 x 25
A = 78.5 cm2

d/2 = r
15/2 = r
15/2= 7.5
A=</sup> π^{r2
A= 3.14 x 7.52}
A= 3.14 x 56.25
A= 176.63 cm2



Cylinder Volume and Volume Problems





a) d/2= r
10/2= 5cm^2

v= (πr^2) * h
v= (3.14 x 5 x 5 ) x 20
V= (78.5) x 20
V= 1570cm^3

b) d/2= r
1/2= o.5m^2

v= (πr^2) x h
v= (3.14 x 0.5 x o.5) x 1
v= (0.785) x 1
v= o.785m^3

c) d/2= r
18/2= 9cm^2

v= (πr^2) x h
v= (3.14 x 9 x 9) x 7.5
v= 254.34 x 7.5
v= 1907.55cm^3

Victoria's Final Percent Post, Cylinder & Volume Problems

A Percent is a number out of 100 or 100% . A hundred grid contains 100 squares, if you fill in 50 squares that is 50%. If you fill in 30 squares that would be 30%

4.1 Representing Percents.

To represent a percent, you can shade in squares on a hundred grid. A completely shaded hundred grid represents 100%. If the percent is greater than 100 you will have to shade more than one grid. To represent a percent between 0% and 1%, shade part of one square from the hundred grid.

4.2 Fractions, Decimals, and Percents

Fractions, decimals, and percents can be used to represent numbers in various situation.Percents can be written as fractions and as decimals.

4.3 Percent of a Number.

You can use strategies like halving, doubling, and dividing by ten to finds of numbers.

To calculate the percent of a number, write the percent as a decimal and then multiply by the number.

4.4 Combining Percents.

Percents can be put together by adding to solve questions.

To calculate the increase in a number, you can add the combined percent amount to the original number. You can multiply the original # by a single percent greater than 100.

a) V=( πxr^2 )x h
V=( 3.14 x 5cm ) x 23
V=
b) V=( πxr^2 )x h
V= ( 3.14 x 14 ) x 12
V=
c) V= (πxr^22 ) x h

Hanna's Volume Scribe Post

18. Suki has 30 small linking cubes.

a) She wants to use 18 of them to make a large cube. Is this possible? Why or why not?

b) What number of linking cubes would she use to construct the largest cube she can
possibly make?


My Answers

a) It is not possble to make a large cube with only 18 linking cube because there isn't a whole number that can make 18 when cubed. To get 18, each side of the cube should have 2.6207413942088964 cubes.


b) v = s x s x s
v =1y x 1y x 1y
v = 1y³

v = s x s x s
v = 2 x 2 x 2
v = 8³

v = s x s x s
v = 3 x 3 x 3
v = 27³

v = s x s x s
v = 4 x 4 x 4
v = 64³

The largest cube Suki can make with only 30 linking cubes uses 27 cubes.


19. Melissa has three glass vases. She wants to use one as a decorative fish tank for Harvey the
guppy. Which will give Harvey the most water to swim in?

Cube Tank

v = s x s x s
v = 7cm x 7cm x 7cm
v= 343cm³

Rectangular Prism Tank

v = l x w x h
v = 10cm x 9cm x 4cm
v = 360cm³

Triangular Prism Tank

v = b x h(1) / 2 x h(2)
v = 7cm x 5cm / 2 x 21cm
v = 367.5cm³

The triangular prism tank can contain the most water.

Cylinder Volume and Volume Problems


a) r = d ÷ 2
r = 2m ÷ 2
r = 1m

v = π x r² x h
v = 3.14 x 1² x 0.6

v = 1.884m³

b) h = v ÷ a of b

h = 1.256 ÷ 3.14

h = 0.4m



c) v = pi x r² x h

v = 3.14 x 1² x 0.2

v = 0.628m³





d ÷ 2

r = 120 ÷ 2
r = 60cm


a) v = pi x r² x h

v = 3.14 x 60² x 18

v = 3391.2cm³




b) v = l x w x h

v = 30 x 22 x 24

v = 15840cm³





c) 15840cm³ x 1 = 15840cm³

15840cm³ x 2 = 31680cm³

15840cm³ x 3 = 47520cm³





She would have to fill it 3 times

Kevin's Volume Scribe Post

Cylinder Volume and Volume problems

7.3


Volume of he cylindrical elements are as follows (If I wrong please correct me!):
Zarya FGB= 166.27m^3
Unity Node = 91.35m^3
Zvezda service module = 181.40m^3
Z1 Struss = 67.85m^3
P6 Truss Solar Array = 307895.18m^3
Destiny = 123.37m^3

A) P6 Truss Solar Array
B) Height = 117.8m
Radius = 16.05m

Volume =










7.4

Arween's Volume Post

Cylinder Volume and Volume Problems
7.3
















Answers:
a)

R=D ÷ 2
R=2 ÷ 2
R=1m

V=π • r • r • h
V=(3.14•1•1)X 0.6
V=1.884m3
The volume of the tub with a depth of 0.6m is 1.884m3

b)

To find the height of the tub divide the Volume to the Area of the Base. Before that find the radius first.

R=D ÷ 2

R=2 ÷ 2

R=1m

Then find the area of the circle.

A=π * r * r

A=3.14 * 1 * 1

A=3.14m

Then Divide the Volume to the area of the circle.

H=1.256m3 ÷ 3.14m

H=0.4m

c)

Step 1. Find the volume of the tub with a depth of 0.5m

Step 2. Find the volume of the tub with a depth of 0.7m

step 3. Subtract the volume of the the tub with a depth of 0.7m to the tub with 0.5m.

Like this......

R=D ÷ 2

R= 2 ÷ 2

R=1m

Tub with a depth of 0.5m:

V=π*r*r*h

V=(3.14*1*1) * 0.5

V=1.57m3

Tub with a depth of 0.7m

V=π*r*r*h

V=(3.14*1*1) * 0.7

V=2.198m3

Then Subtract.

2.198m3 - 1.57m3

=0.628m3

it would take 0.628m3 to fill the tub with water at 0.7m.

7.4

Answer:

Outside:

R=D ÷ 2

R=1 ÷ 2

R=0.5m

V=π * r * r * h

V=(3.14*0.5*0.5) * 10

V=7.85m3

Inside:

R= D ÷ 2

R= 0.8 ÷ 2

R=0.4m

V=π * r * r * h

V=(3.14*0.4*0.4) * 10

V=5.024m3

Now Subtract the outside diameter to the inside diameter.

7.85m3 - 5.024m3

=2.8m3

Cathlene's Volume Post

Chapter 7 Section 7.1 Page 246 to 253

Chapter 7 Section 7.2 Page 246 to 253

Chapter 7 Section 7.3 Page 246 to 253

Chapter 7 Section 7.4 Page 268 to 275

Sam's Volume Post
Cylinder Volume and Volume Problems

Section 7.3

The formula for solving a cylinder problem is π.r.r.h .

the height of the Capture envelope is 10.15
the diameter of the Capture envelope is 23.3

r=d/2
r=23.3/2
r=10.15

v=π.r.r.h
v=3.14x10.15x10.15x10
v=3234.9cm^3

The volume of the Capture Envelope is 3234.9cm^3 !!!!!!

Section 7.4















The formula for a triangular prism is bxh1xh2/2

height of the chocolate bar is 5cm
base of chocolate bar is 5.6cm
length of chocolate bar is 20cm

v=bxh1xh2/2
v=5.6x5x20/2
v=280cm^3
v=280x64
v=17,920cm^3

The volume of the chocolate display is 17,920cm^3 !!!!!!

Errol's Scribe Post

a) V= BxH1/2 . H2
V= 7x7/2 x 12
V= 294cm^3

b) V= BxH1/2 . H2
V= 8.1x2.2/2 x 15
V= 133.65m^3

c) V= BxH1/2 . H2
V= 210x320/2 x 400
V= 13440000mm^3






















She should get the Popcorn Lover's one because:

































Here is how I solved it:

Ishaka's Volume Scribe Post

7.3

d/2=r

20.3/2=r10.15=r (π.r.r)x h=v
(3.14 . 10.15 . 10.15) x 10=v
323.49065cm^2 x 10cm=v
3 234.9065cm^3=v
The Answer: 3 234.9065cm^3

7.4

M.I

* height = 10m
* diameter = 1 (outside) and 0.8 (inside)

Outside : r = d/2 r = 1/2 r = 0.5 v = (π x r x r) x h v = (3.14 x 0.5 x 0.5) x h v = 0.78m^2 x 10m v = 7.8m^3 Inside : r = d/2 r = 0.8/2 r = 0.4 v = (π x r x r) x h v = (3.14 x 0.4 x 0.4) x h v = 0.5m^2 x 10m v = 5m^3 Inside = 5m^3 Outside = 7.8m^3 volume2 - volume1 = volume of culvert 7.8m^3 - 5m^3 = 2.8m^3 The concrete required to make the culvert is 2.8m^3

Carl's Volume Scribe Post

Cylinder Volume and Volume problems

Here are the following examples of questions that I answered :

Section 7.3 : Pages 265-267




















* height = 10cm
* diameter = 20.3cm
r = d/2
r = 20.3/2
r = 10.15cm

v = π x r x r x h
v = 3.14 x 10.15 x 10.15
v = 323.49cm2
v = 323.49 x 10

Final Answer : v = 3234.9cm3

Section 7.4 : Pages 272-275

* height = 10m
* diameter = 0.8 (inside) and 1 (outside)

Inside :
r = d/2
r = 0.8/2
r = 0.4

v = π x r x r x h
v = 3.14 x 0.4 x 0.4 x h
v = 0.5 x 10
v = 5m

Outside :
r = d/2
r = 1/2
r = 0.5

v = π x r x r x h
v = 3.14 x 0.5 x 0.5 x h
v = 0.78 x 10
v = 7.8m

We are trying to find a way to find the volume of the concrete needed to make the culvert. So what I did is that I took the total from the volume of the inside concrete and just subtracted from the total of the outside concrete.

Inside = 5m
Outside = 7.8m

What I did :
7.8 - 5m = 2.8m3

The concrete required to make the culvert is 2.8m3