C1 and C2 DOUBTS HERE!!!!!

[~ Miss Relina ~]

hi i need help in a log question

LOG(base 2)X + LOG (base 4 ) = 2 Answer is 2.52 how?

there's something miissing in this question 

[~ Miss Relina ~]

here we go  
a question to sove plz

click on the image for larger view

anyone answer me plz 

[The Golden Girl =D]

here we go  
a question to sove plz

click on the image for larger view

I've got that Question solved somewhere in my C2 copybook ... I'll upload it tomorrow when I come back from School ' I'm Busy now =,='

Remind me because I tend to forget

[~ Miss Relina ~]

I've got that Question solved somewhere in my C2 copybook ... I'll upload it tomorrow when I come back from School ' I'm Busy now =,='

Remind me because I tend to forget

I've got that simple answerin my C2 book  too which I don't understand and I need steps to be done for me
Thanks in advance

[The Golden Girl =D]

I've got that simple answerin my C2 book  too which I don't understand and I need steps to be done for me
Thanks in advance

Okay I'll see what I kind do tomorrow inshAllah =]

[WARRIOR]

there's something miissing in this question 

yea sorry

LOG (base 2 )X + LOG(base 4) X = 2

i have another question

given that p = log( base q) 16 , express in terms of p

log ( base q) 8Q

this is in C2 , PAGE 48 , 3 , B

[astarmathsandphysics]

LOG (base 2 )X + LOG(base 4) X = 2

you have to change the base from 4 to 2
use log_a x = (log_b x)/(log_b a) so log_4 x = (log_2 x)/(log_2 4)=(log_2 x)/2
then log_2 x +1/2 log?2 x =2
3/2 log_2 x = 2 so log_2 x =4/3 so x =2^(4/3)

2nd question
log_q 8q = log_q 8 +log_q q = log_q 2^3 + 1 = log_q (16^(1/4))^3 + 1 = log_q 16^(3/4) +1 = 3/3 log_q 16 +1 =3/4 p +1

[WARRIOR]

LOG (base 2 )X + LOG(base 4) X = 2

you have to change the base from 4 to 2
use log_a x = (log_b x)/(log_b a) so log_4 x = (log_2 x)/(log_2 4)=(log_2 x)/2
then log_2 x +1/2 log?2 x =2
3/2 log_2 x = 2 so log_2 x =4/3 so x =2^(4/3)

2nd question
log_q 8q = log_q 8 +log_q q = log_q 2^3 + 1 = log_q (16^(1/4))^3 + 1 = log_q 16^(3/4) +1 = 3/3 log_q 16 +1 =3/4 p +1

I undestood them now thanks to you. Thanks a milllion sir !

[astarmathsandphysics]

no probs. See the c2 note on my website too.

[The Golden Girl =D]

here we go  
a question to sove plz

click on the image for larger view

Next time you post a C2 Question make sure you mention which Exercise and which page it's on plus which number it is ,it'll make it much easier for us to answer you 'those who do have the C2 book'.

HERE you go

P = 100 cm
A = A cm²

a) P = 2r + r theta                   => P stand for Perimeter
     100 = 2r + r theta
      r theta = 100 - 2r
      theta = (100/r) - 2

Area = 0.5 * r² * theta
         = 0.5 * r² ( (100/r) - 2 )
           = 0.5 (100 r - 2r²)
                        So,   A = 50r - r²


b) i) r = (-b/2a) = (-50/2(-1)) = 25             So , r = 25

A is Maximum Value because a = -1 < 0     => I don't know what my teacher meant by that ' Sorry I can't remember what he said about this thing =,= '


ii)  theta = (100/25) -2 = 4 - 2 = 2 rad   So , theta = 2 rad

iii)

Area = 50(25) - (25)²
       = 1250 - 625

   So , A = 625 cm²

I'll attach An Image so you understand what my teacher did to the drawing so check Attachment =]

[astarmathsandphysics]

thanks gg

[The Golden Girl =D]

No Problem Sir =]

[MKh]

I don't have the textbook. can you scan the questions

Thanks for replying Sir. I am srry i don't have a scanner. But Kimo Jesus has typed the questions here.

[~ Miss Relina ~]

Next time you post a C2 Question make sure you mention which Exercise and which page it's on plus which number it is ,it'll make it much easier for us to answer you 'those who do have the C2 book'.

HERE you go

P = 100 cm
A = A cm²

a) P = 2r + r theta                   => P stand for Perimeter
     100 = 2r + r theta
      r theta = 100 - 2r
      theta = (100/r) - 2

Area = 0.5 * r² * theta
         = 0.5 * r² ( (100/r) - 2 )
           = 0.5 (100 r - 2r²)
                        So,   A = 50r - r²


b) i) r = (-b/2a) = (-50/2(-1)) = 25             So , r = 25

A is Maximum Value because a = -1 < 0     => I don't know what my teacher meant by that ' Sorry I can't remember what he said about this thing =,= '


ii)  theta = (100/25) -2 = 4 - 2 = 2 rad   So , theta = 2 rad

iii)

Area = 50(25) - (25)²
       = 1250 - 625

   So , A = 625 cm²

I'll attach An Image so you understand what my teacher did to the drawing so check Attachment =]

well thanx alot
 

[~ Miss Relina ~]

here we go those  two questions in read plz  

attached
                  if u have C2 book then its exercise 10D  Q1 H  page 167


Thanks in advance