IMPORTANTTT

[halosh92]

ive got a college exam coming up in 2 days for maths
c1, c2 and probably c3 and c4...
could someone help me out with these questions plz
thxxxx


x^4 - 625

which are the following are its factors
a) x - 5
b) x + 5
c) x^2 + 25

is it a)
 or b)
or c)
 or (a) and (b)  
or all of them ?

[cooldude]

ive got a college exam coming up in 2 days for maths
c1, c2 and probably c3 and c4...
could someone help me out with these questions plz
thxxxx


x^4 - 625

which are the following are its factors
a) x - 5
b) x + 5
c) x^2 + 25

is it a)
 or b)
or c)
 or (a) and (b)  
or all of them ?
 

this is of the form a^2-b^2=(a+b)(a-b)
root x^4=x^2 and root 625=25
therefore (x^2+25)(x^2-25), now we again use the identity
(x^2+25)(x+5)(x-5)
therefore a,b and c are correct

[Alpha]

ALL 3 are correct.

You know the concept: difference of perfect squares?

x^4 - 625 = (x^2)^2 - 5^4
         
                = (x^2)^2 - (5^2)^2

                = (x^2 - 5^2)(x^2 + 5^2)

                = (x - 5)(x + 5)(x^2 + 25)

[halosh92]

thx to both of you <3

[halosh92]

am sorry the questions are basic
but could u gimmi the answer for this

2^40 + 2^40+ 2^40 + 2^40

[cooldude]

am sorry the questions are basic
but could u gimmi the answer for this

2^40 + 2^40+ 2^40 + 2^40

= 4*(2^40)
= (2^2)*(2*40)
now use the law of indices (2^x)*(2^y)=(2^(x+y))
therefore --> 2^(40+2)=2^42

[halosh92]

ok and this one  also

  solve for x : l 5x-6 l   + 3 = 10
the solution set is:
a)-1/5
b)13/5
c)7/5
d)-7/5
e) none of these.

[cooldude]

ok and this one  also

  solve for x : l 5x-6 l   + 3 = 10
the solution set is:
a)-1/5
b)13/5
c)7/5
d)-7/5
e) none of these.

7= | 5x-6 |
7=5x-6 and 7=-(5x-6)
therefore x=13/5, -1/5
therefore a and b

[halosh92]

32^(2/5)  +  16^(1/4)

[cooldude]

32^(2/5)  +  16^(1/4)

(32^(1/5))^2 + (2^4)^(1/4)
2^2 + 2=6

[mousa]

Arent those for AUS maths placement test??? 

[halosh92]

1) if  f(x) = 5x + 1
and g(x) = x^2
then find  g(f(x))=

2) let f(x) = 1/x^2
as x approaches  negative infinity, f(x)
a) approaches 0
b) approaches  negative infinity
c) approaches infinity
d) approaches 1
e) approaches -1

3) given that k>0 solve the exponential equation : e^2t=k   for t
a) (k/2)
b)k^2
c)sqrt k
d) 2k
e) none of these


thxxxxx  

[halosh92]

Arent those for AUS maths placement test??? 

yepppp 
u heading there 2 ?

[cooldude]

1) if  f(x) = 5x + 1
and g(x) = x^2
then find  g(f(x))=

2) let f(x) = 1/x^2
as x approaches  negative infinity, f(x)
a) approaches 0
b) approaches  negative infinity
c) approaches infinity
d) approaches 1
e) approaches -1

3) given that k>0 solve the exponential equation : e^2t=k   for t
a) (k/2)
b)k^2
c)sqrt k
d) 2k
e) none of these


thxxxxx  

1)gfx=(5x+1)^2
=25x^2+10x+1
2) a)approaches 0
1/(-infinity)^2=0

3) 2t=ln k
t=ln k/2

[halosh92]

if sin p = 2/3  and                   pie/2 <= p <= pie                , then tan p is equal to

A) -2/sqrt 5
B)  2 / sqrt 13
c) sqrt 5/2
D)- 1 /sqrt 13
E) 2/sqrt 5