Pages: 1

 

2009-02-09 14:08:48
Oisín | Irishpride
Banned
+118|5591|Dublin, Ireland
I got a maths exam and I need to understand the answers for the following questions, the person who answers all questions with correct method and the proper layout gets 10 karmaz!

#1 - Algerbra

(b)(1) simplify:
(x-√x+1)(x+√x+1) where x+1 > 0

(btw, all of x+1 is under the sq root, i just couldnt type it)

(11) Hence, or otherwise, find the value of x for which
(x-√1+x)(x+√1+x)=11

(c)(1) find the value of x for which
(27 to the power of 1/3)(√27)=3 to the power of 2x+1
(sorry cant type powers)

(11) Find the smallest interger number x such that
x+2 over 3 -1 is less than or equal to 2x+1 over 2
(if that makes sense in writing?)

#2 - Algebra

Q3. (a) given that qp/s = r-s, express q in trms of p,r and s

(qp/s means qp over s lol)

(b) Solve the simultaneous equations
3x+4y/2 = 2x-y (that means 3x+4 all over 2)

3/2x -4y=5

(c) Let f(x) = 6xcubed+ 5xsquared-3x-2

(1) Show that 6xsquared-x-2 is a factor of f(x)
(11) Hence or otherwise slve the equation f(x)=0

#3 - Complex numbers (Imaginary Number)

(a) given that i² = -1 simplify 3+4(3+2i)-i(3+5i)
and write your answer in the form x+yi

(b) let u= -2-5i

(1) Dont gotta do dis 1.

(11) express u+3/u+2i in the form p+qi ...(thats u+3 ovr u+2i in case that look confusing)

(111) Hence, or otherwise, calculate |u+3/u+2i|

(c) (1) solve z²-6z+10=o write your answer in the form a+bi

(11) Let w= 1+i
Find the real numbers k and t such that k(w+i)+w(k-12)=5

Please help me guys!

Last edited by Oisín | Irishpride (2009-02-09 14:12:57)

2009-02-09 14:11:42
ghettoperson
Member
+1,943|6646

Yeah fuck that...
2009-02-09 14:13:12
Oisín | Irishpride
Banned
+118|5591|Dublin, Ireland

ghettoperson wrote:

Yeah fuck that...
10 karmaz!!! PLZ!!!

AT LEAST ANSWER 1 PART IN 1!

Last edited by Oisín | Irishpride (2009-02-09 14:13:36)

2009-02-09 14:16:34
P_Diaz
Member
+6|5558|USA
If I didn't have my Calculus test on wednesday I'd deff help you, but I have no time. I have to studyyy
2009-02-09 14:20:02
Oisín | Irishpride
Banned
+118|5591|Dublin, Ireland
Help me moar?
2009-02-09 14:39:56
Fosterthelegend
:)
+97|6612|Kent, UK

Oisín | Irishpride wrote:

Help me moar?
gimmie kamas
2009-02-09 14:42:37
MAGUIRE93
High Angle Hell
+182|6191|Schofield Barracks
i would help but i am only in algebra 1 in high school. :p
2009-02-10 01:39:20
beerface702
Member
+65|6690|las vegas
get a tutor.

that stuff is hard man
2009-02-10 02:23:25
JahManRed
wank
+646|6625|IRELAND

I hated math all through school and I still do. Your on your own buddy.

Use the Internet to cheat. All the kids are at it.
2009-02-10 02:40:01
Catbox
forgiveness
+505|6713
x=2
Love is the answer
2009-02-10 03:01:47
Hakei
Banned
+295|5992
#3 - Complex numbers (Imaginary Number)

(a) given that i² = -1 simplify 3+4(3+2i)-i(3+5i)
and write your answer in the form x+yi

3 + (12 + 8i) -(3i - 5i²)
3 + 12 + 8i - 3i - -5
3 + 12 +5 + 8i - 3i
20 + 8i - 3i

=20 + 5i

(b) let u= -2-5i

(1) Dont gotta do dis 1.

(11) express u+3/u+2i in the form p+qi ...(thats u+3 ovr u+2i in case that look confusing)
(-2 - 5i) + 3
---------------
(-2 - 5i) + 2i

1 - 5i
------
-2 - 3i

Now what we need to do is, multiply out the bottom by the reciprocal (This is simply the same thing with the opposite sign)

1 - 5i      -2 + 3i
-------  x  -------
-2 - 3i     -2 + 3i

-2 + 3i + 10i - 15i²
----------------------
4 - 6i + 6i -9i²

-2 + 13i -15i²
----------------
     4 - 9i²

-2 + 13i - -15
-----------------
4 - -9

13 + 13i
----------
13

1 + 1i
-     -
1     1

1 + 1i




(111) Hence, or otherwise, calculate |u+3/u+2i|

1+ sqrt-1
(c) (1) solve z²-6z+10=o write your answer in the form a+bi

For this we need the quad formula, which is:

https://www.mathwarehouse.com/quadratic/images/the-quadratic-formula.gif

Hence we have...

-6z +- sqrt(36z² - 4x1x10)
--------------------------------
                    2
-6z +- sqrt(36z² - 40)
------------------------------
                2

-3z +- sqrt (36z² - 40)
          -------------------
                     2

As far as I know, that is sound enough for an answer, as it is in the form a + bi.


(11) Let w= 1+i
Find the real numbers k and t such that k(w+i)+w(k-12)=5

k(1+i+i) + (1+i)(k-12) = 5

k + 2ki + (k - 12 + ki - 12i) = 5

2k + 3ki -12 - 12i = 5

2k + 3ki -17 - 12i = 0

2k + 3ki - 12i = 17

Question doesn't make sense, seeing as you haven't included T into the formula :p, but the above working is what you need to do.

At the end of the day, all you need to show is an understanding of how the working is done, the final answer isn't important, in most tests it's demonstrating the ability to solve, not the bottom line.


The other stuff I'll try write on in the morning, but right now I have to work on some unit circle questions, hit me on xfire if you need help.
2009-02-10 04:23:53
Dilbert_X
The X stands for
+1,810|6103|eXtreme to the maX
It would take a bit longer than 24 hrs to teach you all that maths.
It would take me at least that long just to remember it.
Русский военный корабль, иди на хуй!
2009-02-10 07:59:17
Jussimies
Finnish commander whore
+76|6581|Finland
Here are the answers to the first two calculations:
https://img152.imageshack.us/img152/1180/nimzk9.th.jpg
2009-02-10 08:45:15
DrunkFace
Germans did 911
+427|6678|Disaster Free Zone
#1 - Algerbra
(b)(1)
=x2-(√(x+1))2
=x2-x-1 for x=> -1

(11)
x2-x-1 = 11
=x2-x-12 = 0
x = (1±7)/2
x = -3, 4 (because x=> -1)

#2 - Algebra
Q3 (a)
qp/s = r-s
qp = s(r-s)
q = (sr-s2)/p  (where s,p =/= 0)

(b)
(3x+4y)/2 = 2x-y (1)
3/2x -4y=5 (2)

3x+4y = 4x-2y
6y = x
sub (1) into (2)
3/2(6y) - 4y = 5
9y-4y = 5
5y = 5
y=1
x=6

(c)
f(x) = 6x3+5x2-3x-2
(1) 6x2-x-2
bah, use long division... it's not hard just forgotten how its been to long.

Last edited by DrunkFace (2009-02-10 08:51:36)

2009-02-10 09:17:38
Oisín | Irishpride
Banned
+118|5591|Dublin, Ireland

Hakei wrote:

#3 - Complex numbers (Imaginary Number)

(a) given that i² = -1 simplify 3+4(3+2i)-i(3+5i)
and write your answer in the form x+yi

3 + (12 + 8i) -(3i - 5i²)
3 + 12 + 8i - 3i - -5
3 + 12 +5 + 8i - 3i
20 + 8i - 3i

=20 + 5i

(b) let u= -2-5i

(1) Dont gotta do dis 1.

(11) express u+3/u+2i in the form p+qi ...(thats u+3 ovr u+2i in case that look confusing)
(-2 - 5i) + 3
---------------
(-2 - 5i) + 2i

1 - 5i
------
-2 - 3i

Now what we need to do is, multiply out the bottom by the reciprocal (This is simply the same thing with the opposite sign)

1 - 5i      -2 + 3i
-------  x  -------
-2 - 3i     -2 + 3i

-2 + 3i + 10i - 15i²
----------------------
4 - 6i + 6i -9i²

-2 + 13i -15i²
----------------
     4 - 9i²

-2 + 13i - -15
-----------------
4 - -9

13 + 13i
----------
13

1 + 1i
-     -
1     1

1 + 1i




(111) Hence, or otherwise, calculate |u+3/u+2i|

1+ sqrt-1
(c) (1) solve z²-6z+10=o write your answer in the form a+bi

For this we need the quad formula, which is:

http://www.mathwarehouse.com/quadratic/ … ormula.gif

Hence we have...

-6z +- sqrt(36z² - 4x1x10)
--------------------------------
                    2
-6z +- sqrt(36z² - 40)
------------------------------
                2

-3z +- sqrt (36z² - 40)
          -------------------
                     2

As far as I know, that is sound enough for an answer, as it is in the form a + bi.


(11) Let w= 1+i
Find the real numbers k and t such that k(w+i)+w(k-12)=5

k(1+i+i) + (1+i)(k-12) = 5

k + 2ki + (k - 12 + ki - 12i) = 5

2k + 3ki -12 - 12i = 5

2k + 3ki -17 - 12i = 0

2k + 3ki - 12i = 17

Question doesn't make sense, seeing as you haven't included T into the formula :p, but the above working is what you need to do.

At the end of the day, all you need to show is an understanding of how the working is done, the final answer isn't important, in most tests it's demonstrating the ability to solve, not the bottom line.


The other stuff I'll try write on in the morning, but right now I have to work on some unit circle questions, hit me on xfire if you need help.
GG WP

I had my exam 4 hours ago

Nvm now, thx do mate <3

 

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