Questions   
Sort: 
 #3
avatar+33661 
+10
Jun 15, 2014
 #16
avatar+11912 
+3
Jun 15, 2014
 #2
avatar+11912 
0
Jun 15, 2014
 #2
avatar+118723 
+10

Hi Jedithious,

It is always great to see you in the forum!  

 

1) The ratio of the surface areas of two cubes is 64:81. What is the ratio of the edges of the cubes?

the answer will just be sqrt(64):sqrt(81) = 8:9

Now I will show you why

$$\dfrac{6l_1^2}{6l_2^2}=\frac{64}{81}\qquad so \qquad \dfrac{l_1^2}{l_2^2}=\frac{64}{81}\qquad so \qquad \dfrac{l_1}{l_2}=\frac{8}{9}$$

 

2) If the ratio of the volumes of two similar cylinders is 125:27, what is the ratio of their surface areas?

volume is U3,  Surface area is U2

so ratio of surface areas will be  125^(2/3):27^(2/3) = 25:9

 

3) A model of a rocket was built to a scale of 1:40. 

a) The length of the model was 75 cm. What was the length of the actual rocket, in metres?

75*40=3000cm=30m

b) The area of metal used to cover the curved surfaces of the model was 2750 cm2. What area of metal covered the curved suraces of the actual rocket, in square metres?

1:1600 = 2750:2750*1600=4,400,000cm2=440m2     (Little edit here)

c) The nose cone of the rocket had a volume of 96 m3. What was the volume of the nose cone of the model, in cubic centimetres?

1:403 = x:96  

$$\frac{1}{64000}=\frac{x}{96}\\\\
96\times \frac{1}{64000}=x\\\\
x=1.5\times 10^{-3}\quad m^3$$

100cm in a metre

1003cm3 in a m3

$$x=1.5\times 10^{-3}\times 10^6 = 1.5\times 10^{3} = 1500cm^3$$

-------------------------------------------------------------

NOW I have taken shourt cuts (maybe I am not allowed to) and my answers are different from CPhill if I get time I will check the long way but I think one of the other mathematicians will validate one of our answers before then. 

Jun 15, 2014
 #1
avatar+3502 
+8
Jun 15, 2014
 #6
avatar+118723 
0

I use this method all  the time!  It is the BEST way to do it!

Factoring trinomials using the grouping method.  

This is a really good clip. 

https://www.youtube.com/watch?v=HvBiJ9W00Z4

If you have any questions just ask.

--------------------------------------------------

(I'd like to point out to other answerers that this is just one of the addresses included in my 

"Information pages worth Keeping and Developing" thread - It has been there for a while!

Jun 15, 2014

1 Online Users

avatar