STENON:
thanks, STENON is in the mood of her final exams.
Good morning,
Hope you are good?

Morning stenon, I am good. I wish you a resounding succes.

naturalwaves:

Morning stenon, I am good. I wish you a resounding succes.

. Thanks...I also wish you the best in your endeavors.

I greet everyone in the house.

sorry 4 bringing up something too elementary,pride goes b4 a fall,lemme embarass myself here than in a real situation
1)explain 0^0=1
2ss1 maths (2x)^1/2 × (2x^3)^3/2

Following:
sorry 4 bringing up something too elementary,pride goes b4 a fall,lemme embarass myself here than in a real situation
1)explain 0^0=1
2ss1 maths (2x)^1/2 × (2x^3)^3/2

1.

Let 0⁰=x
Log₂0⁰ = Log₂x
0 x Log₂0 = Log₂x
but 0 x Log₂0 = 0

Log₂x = 0
x = 2⁰ = 1


2.

(2x)^1/2 × (2x^3)^3/2

(2)^(1/2).x^(1/2) . 2^(3/2). x^(9/2)
2^(1/2)+(3/2).x^(1/2)+(9/2)
2⁽²⁾.x⁵
4x⁵

Antoinne:
1.
Let 0⁰=x Log₂0⁰ = Log₂x 0 x Log₂0 = Log₂x but 0 x Log₂0 = 0
Log₂x = 0 x = 2⁰ = 1

2.
(2x)^1/2 × (2x^3)^3/2
(2)^(1/2).x^(1/2) . 2^(3/2). x^(9/2) 2^(1/2)+(3/2).x^(1/2)+(9/2) 2⁽²⁾.x⁵ 4x⁵

thanks a lot my oga,...make una no blame me o,I don suffer 4 dt question hand

Plz guys, integrate x^x wrt x

timonski:
Plz guys, integrate x^x wrt x

if I solve am,how much you go give me

Antoinne:

1.
Let 0⁰=x
Log₂0⁰ = Log₂x
0 x Log₂0 = Log₂x
but 0 x Log₂0 = 0
Log₂x = 0
x = 2⁰ = 1
2.
(2x)^1/2 × (2x^3)^3/2
(2)^(1/2).x^(1/2) . 2^(3/2). x^(9/2)
2^(1/2)+(3/2).x^(1/2)+(9/2)
2⁽²⁾.x⁵
4x⁵

The first question is wrong. 0^0 can NEVER be 1. That's why they say any nonzero number raise to 0 is one. Try using ur calculator 0^0

Following:
sorry 4 bringing up something too elementary,pride goes b4 a fall,lemme embarass myself here than in a real situation
1)explain 0^0=1
2ss1 maths (2x)^1/2 × (2x^3)^3/2

The first question is wrong. 0^0 can NEVER be 1. That's why they say any nonzero number raise to 0 is one. Try using ur calculator 0^0

hilaomo:
if I solve am,how much you go give me

hahaahaaaaha. pls solve 4 him na

masperano:


The first question is wrong. 0^0 can NEVER be 1. That's why they say any nonzero number raise to 0 is one. Try using ur calculator 0^0

hi. i just used my calc nd it says "1"

ikoyila:
hi. i just used my calc nd it says "1"

Functionality Error the right answer should be indeterminate. You may use a non scientific calculator to confirm it

Try this. Find dy /dx of. (√x)^-(√y)

Pls, someone should help me solve this.

Express as single fraction

(x-y) / (x^1/3 - y^1/3)

Thank you.
Cc: agentofchange1

Umartins1:
Pls, someone should help me solve this.




Thank you.

Cc: agentofchange1

apply difference of two cubes ..then simplify

by defn,

f^3 -g^3 = (f-g)(f^2 +fg +g^2)

now

put x=(x^1/3)^3 =f & y=(y^1/3)^3 = g

thus we get x^2/3 +(xy)^1/3 +y^2/3


guess that's it.

agentofchange1:


apply difference of two cubes ..then simply

If I know it, I won't call on u.

Umartins1:


If I know it, I won't call on u.

^^ modified
check above

Umartins1:
Pls, someone should help me solve this.



Thank you.
Cc: agentofchange1

(x-y) / (x 1/3 - y 1/3 ) = (x-y)÷(x-y)^1/3 = (x-y)^1-1/3 = (x-y)^2/3 or √(x-y)^3

pls can someone send me a link I can download advanced engineering mathematics by hk dass
please.

ikoyila:
(x-y) / (x 1/3 - y 1/3 ) = (x-y)÷(x-y)^1/3
= (x-y)^1-1/3
= (x-y)^2/3 or √(x-y)^3

Chairman,,!!


27^(1/3) - 8^(1/3) = 3-2=1

& (27- 8.)^1/3 = (19)^1/3

so why am I not getting the same thing ?....


sorry bro , your solution is wrong !!

(x^1/3 - y^1/3 ) =/=
(x-y)^1/3


happy learning ...

#greetings to you all.....happy Sunday...

agentofchange1:


Chairman,,!!


27^(1/3) - 8^(1/3) = 3-2=1

& (27- 8.)^1/3 = (19)^1/3

so why am I not getting the same thing ?....


sorry bro , your solution is wrong !!

(x^1/3 - y^1/3 ) =/=
(x-y)^1/3


happy learning ...

#greetings to you all.....happy Sunday...

clearify all dis ur calculations cos its not d same wit my final answer " (x-y)^2/3" OR "√(x-y)^3". i believe u trying to compare my answer wit numerals. d expression u usin 2 compare is not my answer nd thus wrong.


ur answer: x^2/3 +(xy)^1/3 +y^2/3 i dnt tink it answers d question wch says express as a "single fraction". cheeerss..

ikoyila:

clearify all dis ur calculations cos its not d same wit my final answer " (x-y)^2/3" OR "√(x-y)^3". i believe u trying to compare my answer wit numerals. d expression u usin 2 compare is not my answer nd thus wrong.


ur answer: x^2/3 +(xy)^1/3 +y^2/3 i dnt tink it answers d question wch says express as a "single fraction". cheeerss..

hmm..haba my bro, this isn't something we should be arguing on.

every mathematical expression must be testable & hold for all basic mathematical axioms .


A fraction is the ratio of at least two real numbers (say a/b. , for b=/=0 )

if your answer is correct , irrespective of the number we input to test , it'll still hold .

Now, my answer is still a single fraction , (since x =x/1 )

check: let's still use x=27 & y= 8 ( I used them , because they're perfect cubes , you could use others )

I got x^2/3 + (xy)^1/3 +y^2/3 which we can write as

[ (x^1/3 )^2 + x^1/3 * y^1/3 +(y^1/3) ^2 ]/1

for x=27 , y=8

=> (27^1/3)^2 +27^1/3 * 8^1/3 + (8^1/3)^2

=3^2 +(3*2 ) +2^2

= 9+6+4=19

now the original expression was (x-y)/(x^1/3. -y^1/3 )

thus =( 27 - 8. ) / (27^1/3. - 8^1/3 )

= 19/(3-2)=19 /1 =19 .


hence , my result holds for all x,y € Z

Q.E D .


We are all learning bro.

agentofchange1:

hmm..haba my bro, this isn't something we should be arguing on.
every mathematical expression must be testable & hold for all basic mathematical axioms .
A fraction is the ratio of at least two real numbers (say a/b. , for b=/=0 )
if your answer is correct , irrespective of the number we input to test , it'll still hold .
Now, my answer is still a single fraction , (since x =x/1 )
check: let's still use x=27 & y= 8 ( I used them , because they're perfect cubes , you could use others )
I got x^2/3 + (xy)^1/3 +y^2/3 which we can write as
[ (x^1/3 )^2 + x^1/3 * y^1/3 +(y^1/3) ^2 ]/1
for x=27 , y=8
=> (27^1/3)^2 +27^1/3 * 8^1/3 + (8^1/3)^2
=3^2 +(3*2 ) +2^2
= 9+6+4=19
now the original expression was (x-y)/(x^1/3. -y^1/3 )
thus =( 27 - 8. ) / (27^1/3. - 8^1/3 )
= 19/(3-2)=19 /1 =19 .
hence , my result holds for all x,y € Z
Q.E D .
We are all learning bro.

oooohk! nd am a gurl not a bro.
so ur answer is a single fraction, i am truely learning, like criusli..

ikoyila:


oooohk! nd am a gurl not a bro.
so ur answer is a single fraction, i am truely learning, like criusli..

oh..sorry , its OK my dear ...

happy learning/solving. . .


#1luv.

masperano:


The first question is wrong. 0^0 can NEVER be 1. That's why they say any nonzero number raise to 0 is one. Try using ur calculator 0^0

You are only partly right. For a function x^y, with x,y->0, that function will tend to 1. Just imagine 0.000000000001⁰. The answer is 1.
While that function may not be definable at the origin, it doesn't mean you can't assign 1 to it, especially as the limit actually tends to 1. There's a lot of mathematical argument about this, but technically I'm not wrong.

Gurus in. D aous pls hlp o.... Simplify x^2-1/x^3-2x^2-x-2. (A) 1/x+2 (B) x+1/x+2 (C) x+1/x-2 (D) 1/x-2

2. Which of the following binary operation is commutative in the set of integers? (A) a-b=a-2b (B) a.b=a+b-ab (C) a-b=a^2+b (D) a.b=a(a+1)/2.
In triangle XYZ, <YXZ=44, and <XYZ=112, calculate the acute angle between the internal bisectors of <XYZ and <XZY. (A) 12 (B) 56 (C) 68 (D) 78



11. Two perpendicular lines PQ and QR intersect at (1, -1), if the equation of PQ is x-2y+4=0, find the equation of QR. (A) x-2+1=0 (B) 2x+y-3=0 (C) x-2y-3=0 (D) 2x+y-1=0.
Find the area bounded by the curve y=3x^2-2x +1, the ordinate x=1 and x=3 and the x-axis. (A) 24 (B) 22 (C) 21 (D) 20

33. Two variables x and y are such that dy/dx=4x-3 and y=5 when x=2. Find y in terms of x (A) 2x^2-3x +5 (B) 2x^2-3x+3 (C) 2x^2-3x (D) 4.
What is the solution set for the equation /2x-3/=13? (A) (5, (B) (-5,

Is there anyone here that can explain FIXED POINT ITERATION METHOD?

Richiez, stenon and allof u guyz holding stead here, i say welldone Godbless you for me.








Please Guyz seriously i dont wana barge into your thread but voting is presently goin on for the MR NAIRALAND competition. He came 3rd out of 24 contestants in the elimination round1 yesterday. Please click on the link on my signature and vote for my candidate SCENTMARLC alone in the elimination round2 today. Please help your girl nah **sobbing** wont you?

Good afternoon guys. I hail all the gurus in da house.

Pls help me solve this,

if 1atm = 760mmhg = 101325N/m^2

then 1mmhg = XN/m^2

what is X?

Geofavor:
Good afternoon guys. I hail all the gurus in da house.

Pls help me solve this,

if 1atm = 760mmhg = 101325N/m^2

then 1mmhg = XN/m^2

what is X?

X = 101325Nm^-2/760mmHg
= 133.32Nm^-2

dejt4u:
X = 101325Nm^-2/760mmHg = 133.32Nm^-2

thnx. But pls can you explain why you did it like this? I don't understand. Thank you