₦airaland Forum

benbuks: @ bolded is it sec.tita or 2 sec.tita ....?

oops its 2sec tita
thanks

find the values for which the following are TRUE
(a)tan^2¢+2ec¢=5 ........0<=¢<=180
(b)sin¢-cos2¢+3=0........-180<=¢<=180

pls gurus in the house come to my rescue

Davijunson: Wow..... Goes to show so much research was made before the movie TITANIC was shot...the dinning shown here is a replica of what was used in the movie. It's cool indeed. Really cool.

yea not like our own hungrywood where they keep on churning out movies without research

where has he been all these while. oh I forgot 2015 is around the corner

benbuks: ok...you"ll have to go back to your basics of complex numbers

rationally , negative roots are invalid e.g ± √-1 , ±√-2 ,±√-3 e.t c.

solving x² + 1 =0 ..........(*)

produces complex roots , however mathematicians have worked severally on such irrational roots , which later discovered & agreed that solution to say (*) =±√-1

we now use ' i ' or ' j ' = ±√-1 ..........(**)

hence solution of (*) ,x = ± i

having believed that x=± i

=>i² =-1 .......( from (**) )

i³ = i(i²) =i(-1) =-i

i⁴ = (i² ) (i² ) = (-1)(-1)=1

i⁵ = (i⁴ ) i = (1)i = i

we could observe that powers of ' i ' in .

2 ,6 , 10 ,14 . . . =-1

3 ,7 , 11 ,15 . . . =-i

4 , 8 , 12 , 16 . . . = 1

5 , 9, 13 , 17 , . . . =i

& their multiples in that order , gives same result .
.

thanks

benbuks: --------SOLUTION--------------
(a)

= 3(i¹⁰⁴ )i + 5(i⁷² )i -8(i² )²⁵ -2(i¹¹⁸ ) i

=>3i(1) +5i(1) - 8(-1) -2i(-1)

= 10i +8

(b)

by difference of two squares.
we have
[(3i-5)² ]² - [(2-i)² ]²

=>(9i² -30i+25)² -(4-4i+i²)²

=>(16-30i)² - (3-4i)²

=> (256-960i +900i² )-(9-24i+16i² )
=>-544-960i +7 +24i

hence we have ,

-537 -36i

pls I don't get the (a) part, what happened to the powers.
thanks

pls gurus in the house should help with this:
write in standard form, (a)3i^105+5i^73-8i^50-2i^119
(b) (3i-5)^4 -(2-I)^4
PLS I will appreciate if the solutions are accompanied with explanation

efficiencie: Let the matrix A with element a(ij) be defined as:
A=[row(1)(a(11) a(12) a(13) ...a(1m), row(2)(a(21) a(22) a(23) ...a(2m), ... row(n)(a(n1) a(n2) a(n3) ...a(nm)]

Hence if d determinant:
|row(1)(x y z), row(2)(a b c), row(3)(1 2 3)|=5
Then
|row(1)(1 2 9), row(2)(a b 3c), row(3)(x y 3z)|
Multiple of a column
=3. |row(1)(1 2 3), row(2)(a b c), row(3)(x y z)|
Interchange of rows
=-3. |row(1)(x y z), row(2)(a b c), row(3)(1 2 3)|
=-3(5)
=-15

hence
|row(1)(1 2 9), row(2)(a b 3c), row(3)(x y 3z)|=-15

I hope i'm ryt!!!

pls explain better, i'm preparing for an exam

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let's also do what the boys do

if|x y z |
| a b c | =5
| 1 2 3 |
use properties of determinants to find
|1 2 9. |
|a b 3c |
|x y 3z |
pls I need the solution with explanations

Dbestmax: English F9
Mathematics C6
Physics E8
Chemistry E8
Economics F9
Agriculture A1
Biology E8
Geography F9
Igbo Language A1

Not my fault I was sick during d exams...Damn it

with this kind of result na to become herbalist go sure pass

only 5,8 and 22 meaning my debt is just $8.5.
op you shud have added things like,stealing meat frm the pot,stealing mama chukwudi's biscuit and.....

If we are not careful we will wake up one day just to hear that dangote has bought the whole of naijariya

benbuks: my boss @ dej4u has justice already...BUT....

given ax² +bx + c =0

by definition ,

the sum of roots , A+B = -b/a

the product of roots , AB =c/a

but here we are given new roots build a quadratic equation with ,

A-2B & B-2A

now for sum , we have (A-2B) +(B-2A) =A-2B+B-2A
= (A+B) -2(A+B)

recall that A+B =- b/a
=> -b/a -2(-b/a)
= (-b+2b)/a = b/a

& product , (A-2B)(B-2A) =AB-2A² -2B² +4AB

=>

5AB -2(A² + B²)

but A² +B² =(A+B)² -2AB

=> 5AB -2[ ( A+B)² -2AB ]

=> 5c/a -2[ (b/a)² -2c/a]

=> 5c/a -2b² /a² + 4c/a

=> 9c/a - 2(b/a)²

using the general model for building quadratic equations , given roots ,

x² - (sum of roots ) x + product of roots =0

=> x² -(b/a)x +[ 9c/a -2(b/a)² ] =0

simplify further …

hope you get .?

(ax)^2-(ab)x+(9ac-2b^2)
hope I am right

dejt4u: the roots of the given eqn are, A nd B,
then, sum of the root, A+B=-b/a,
Product, AB= c/a,

for the required eqn,
roots are (A-2B) and (B-2A),
sum = (A-2B)+(B-2A) = -(A+B) = -(-b/a) = b/a,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=[s]9(c/a) -2(-b/a) = (9c+2b)/a..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9c+2b)/a = 0,
ax² - bx + (9c+2b) = 0[/s]

modifications.,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=9(c/a) -2(-b/a)² = (9ac-2b)/a²..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9ac-2b)/a² = 0,
multiplying all by a²
(ax)² - abx + (9ac-2b) = 0...QED
sorry for the previous mistake.. Tnx to oga benbuks

pls I thought the last part is supposed to be. (ax)^2-(ab)x+(9ac-2b^2)

safarigirl: haba! Ghana has one bronze medal now, dem never try?

maybe we will dash them some medals after the tournament

south Africa, Nigeria and Kenya doing Africa proud. I wonder if the likes of ghana just went there to answer present

dejt4u: the roots of the given eqn are, A nd B,
then, sum of the root, A+B=-b/a,
Product, AB= c/a,

for the required eqn,
roots are (A-2B) and (B-2A),
sum = (A-2B)+(B-2A) = -(A+B) = -(-b/a) = b/a,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=9(c/a) -2(-b/a) = (9c+2b)/a..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9c+2b)/a = 0,
ax² - bx + (9c+2b) = 0 ...QED

pls explain

efficiencie: Given LOGaA•LOGbB=LOGbA•LOGaB

Starting frm d RHS:
LOGaA•LOGbB
=(LogA/Loga).(LogB/Logb)
=(LogA/Logb).(LogB/Loga)
=LOGbA•LOGaB

Which is the LHS

Nb LOGxX=LogX/Logx

a million thanks and God bless

pls someone should help me with the solution and explanation to this question :
If. A and B are the roots of ax^2+bx+C=o form the equation whose roots are: A-2B, B-2A
PLS note that A and B are alpha and beta respectively.

pls gurus in the house assist me in solving this:
prove that LOGaA•LOGbB=LOGbA•LOGaB
please note that the small letters are the bases of the individual logarithms.

.

weight lifting all the time,aren't there other sports they could win

I believe this list was compiled in alphabetical order,though it still wrong.

They had better stop those bastards or are they keeping quiet because of the $1m loan

kobonaire: NEPA has carry light for my area ....
I was prepared ..... with my generator and fuel

same here ooo,

like if you think only one goal will be scored in this match

goldendiamond: guess you are just joining this thread(after the data loss).welcome on board. whats the business idea. you can post it here but if it will derail the thread feel free to send me an email: ideas2wealth@yahoo.com. you will excel and succeed

just came across thread recently,I will prefer contacting you via email