Post by: halosh92 on August 06, 2010, 05:36:22 pm
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 ?
Post by: cooldude on August 06, 2010, 05:39:38 pm
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
Post by: Alpha on August 06, 2010, 05:46:50 pm
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)
Post by: halosh92 on August 06, 2010, 06:38:06 pm
Post by: halosh92 on August 06, 2010, 07:17:00 pm
but could u gimmi the answer for this
2^40 + 2^40+ 2^40 + 2^40
???
Post by: cooldude on August 06, 2010, 07:27:09 pm
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
Post by: halosh92 on August 06, 2010, 07:32:07 pm
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.
Post by: cooldude on August 06, 2010, 07:39:23 pm
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
Post by: halosh92 on August 06, 2010, 07:50:03 pm
Post by: cooldude on August 06, 2010, 07:52:25 pm
32^(2/5) + 16^(1/4)
(32^(1/5))^2 + (2^4)^(1/4)
2^2 + 2=6
Post by: mousa on August 07, 2010, 08:32:26 am
Post by: halosh92 on August 07, 2010, 09:41:22 am
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 ;D
Post by: halosh92 on August 07, 2010, 09:42:23 am
Arent those for AUS maths placement test??? ::)
yepppp ;D ;D ;D
u heading there 2 ?
Post by: cooldude on August 07, 2010, 10:40:03 am
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 ;D
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
Post by: halosh92 on August 07, 2010, 11:37:18 am
A) -2/sqrt 5
B) 2 / sqrt 13
c) sqrt 5/2
D)- 1 /sqrt 13
E) 2/sqrt 5
Post by: Saladin on August 07, 2010, 11:47:41 am
Post by: Saladin on August 07, 2010, 11:53:10 am
Post by: cooldude on August 07, 2010, 05:18:27 pm
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
sin p=2/3
(sin p)^2+(cos p)^2=1
cos p=root 1-(4/9)=root 5/9
tan p=sin p/cos p=2/3 / (root 5)/3
=2/root 5 therefore e
Post by: mousa on August 07, 2010, 05:25:53 pm
yepppp ;D ;D ;D
u heading there 2 ?
I might. I have done all placement tests.Though i applied for medicine aswell, but still not confirmed till the results are out!!!
By the way, when is your placement test??
Post by: halosh92 on August 08, 2010, 07:40:47 pm
one more question:
cos( 5pie/4) is equal to
a) -1
b) 1/2
c) -1/2
d) -1/sqrt2
e) none of these
is it e ???
another question:
find all the solutions of the equation sin(2x)= 1
given that x is in the interval (0, 2 pie)\
a) pie/2 only
b) pie/4 only
c) pie/4 and 3pie/4
d) pie/4 and 5pie/4
e) none of these.
thxxxx cool dude By the way xD
Post by: Saladin on August 08, 2010, 09:10:16 pm
The answer is (e).
For the second question:
The only answer is (c).
Remembering the sin wave, by making the equation
So, the two answers are:
The working without the explanation is as follows:
Therefore,
45 degrees is equal to
Hope this clears all doubts.
Post by: halosh92 on August 09, 2010, 06:07:19 am
For the first Question:
The answer is (e).basically means that the value is a negative version of cos(
). Because up till
the values for cos are positive, as it is in the ALL quadrant. However, this does not lie in that quadrant, so it is negative version of
.
For the second question:
The only answer is (c).or
(I like working in degrees)
Remembering the sin wave, by making the equation, you put two sin waves into the 0 to 360 gap. Therefore, where 360 degrees was, that position is 180 degrees. Therefore, if we repeat the wave, we repeat the values. Therefore
degrees is also an answer.
So, the two answers are:
The working without the explanation is as follows:
Therefore,
45 degrees is equal to
Hope this clears all doubts.
thankyouuuu yes very clear :D
Post by: Saladin on August 09, 2010, 02:05:44 pm
thankyouuuu yes very clear :D
Glad to be of service! :)