₦airaland Forum

jackpot: bro, are you really interested in the solution or in speed?


Na exam be this?

give the result of the missing numbers ( happy)

jackpot: bro, are you really interested in the solution or in speed?


Na exam be this?

no na brain stroming

simplify (9-0)(9-1)(9-3)(9-4)...(9-n)

find the missing number s; 5,_,1205,_271205,_ ok madam jackpot give me the solutions na

jackpot: Easy one. . .

But there are many answers to this question sha.

how many do you need? I am serious.


NOTE
I know that the sequence is neither an AP or a GP.

ok @ jack pot quickly give me the solution to the problem within 5mins

HINT TO THE PROBLEM 6"X-31(2"X)+32 EQUAL ZERO IS USE LINEAR APPROXIMATION OF BINOMIAL EXPANSION. "*( MEN NA EATING TINX OOOOOOOO)

Alpha Maximus: don't you mean TRANSCENDENTIAL EQUATION?

yes sir is it :HINT USE LINEAR APPROXIMATION

GIVEN THE sequence 5,_,1205,_,271205,_. find the missing numbers . can some one give me the solution to the sequence., ("lol)

jackpot: I presumed he wanted to write 15, 112, 209.

YES

[ I disagree with the third sequence .( it not an AP) THE sequence 15, 112,127

jackpot: You said infinitely more? Can you prove it?

what kind of sequence is this?

Alpha Maximus: can you paraphrase ur question using the appropriate notations like '^'....i.e: 2^x

sorry sir my phone did not have power sign ok permit me to use the symbol ". ok here is the question : 6"x -31(2"x)+32 equal to zero . THIS IS CALLED A TRANSIDENTAL EQUATION (:-hint use LINEAR APPROXIMATION)

-if x is equalto (9-0)(9-1)(9-2)(9-3)...(9-n) for all n is an integer x is ? @ jackpot please kindly give me the solution to the above question with in 10 min

solve for x : given that 6 raise to power x -31(2 raise to power x) +32 equal to zero ( :-hint use linear approximation)

benbuks: X^4 + y^4 =82 .....(a)
x+y = 4 .....(b)
solve

(N100 mtn) first to post correct solution and final answer b4
27mins
show full workings , else nothing for you

*ans 3 * and 1

hardedeji1: use ur "linear approximation"
(1 + x)^n = 1 + nx

6^x - 31(2^x) + 32 = 0
(1 + 5)^x - 31((1 + 1)^x) + 32 = 0
1 + 5x - 31(1 + x) + 32 = 0
26x = 2
x = 1/13

if ur ansa doesn't give a close value to zero.... u can further try d binomial expansion...
i.e
(1 + x)^n = 1 + nx + [n(n - 1)x²]/x!

oga x is 3 .not 1/13

rhydex 247: thanks for dat. d error is frm typing

find dy/dx if f(x) is equal to x raise to power x up to infinity

hardedeji1: use ur "linear approximation"
(1 + x)^n = 1 + nx

6^x - 31(2^x) + 32 = 0
(1 + 5)^x - 31((1 + 1)^x) + 32 = 0
1 + 5x - 31(1 + x) + 32 = 0
26x = 2
x = 1/13

if ur ansa doesn't give a close value to zero.... u can further try d binomial expansion...
i.e
(1 + x)^n = 1 + nx + [n(n - 1)x²]/x!

NOTE PLEASE APPLYING LINEAR APPROX U NEED TO BE CARE FUL clearly, we know that (1+x)raise to power n is equal to (1+nx) for x is a small value like delta . example;: if we have 3 raise to power x using linear approx we have (4-1)raise to power x so factorize 4 we have( 4(1+1/4)) raise to power x clearly compare (1+x) and (1+1/4) you observe that x is 0.25 by comparison ( FOR ALL X IS SMALL)

hardedeji1: 2, 8, 10, 20, 40..

GOOD

find the missing numbers : 2,_10,_40 ( hint using commutative law of addition)

Calculusf(x):
...{logx(base4)}^2=logx(base2)*logx(basea)...{logx(base4)}^2=logx(base4)*logx(base4)...then.logx(base4)*logx(base4)=logx(base2)*logx(basea).....logx(basea)=logx(base4)*logx(base4)/logx(base2)...(i)...{logx(base4)/logx(base2)=1/2}...logx(basea)=1/2.logx(base4)....logx(basea)=logx(base16)...comparing lhs and rhs...a=16

GOOD ! IT CAN ALSO BE OF THE FORM a raise to power log x (base 16) which is equalvent as x . so clearly a is 16

maths guru please what is the solution to the equation 6"x-31(2"x)+32 equal to zero . (.please note the symbol " means raise to power) ie 6 raise to power x -31(2 raise to power x) +32 equal to zero

maths guru please what is the solution to the equation 6"x-31(2"x)+32 equal to zero . (.please note the symbol " means raise to power)

hardedeji1: can anyone help with the solution to this?

3^x + 4^x = 5^x

OBIVIOUSLY 4-1 -is 3 .

we know that: (4-1)raise to power x + (4)raise to power x equals (4+1 ) raise to power x
clearly WE HAVE( 4(1-1/4))raise to power x + (4)raise to power x equal to (4(1+1/4) raise to power x . by linear approximation (1+x) raise to power n we have 1+


nx . when x is very small ( less than 1) clearly we have

4raise to power x( 1-x/4) + 4raise to power x equal to 4 raise to power x(1+x/4) so clearly we obtain the algebraic equation : 1-x/4 +1 is equal to 1+x/4 clearly we obtain
2-x/4 is equal to 1 + x/4
collect like term we have x equal to 2

aysuccess99: i hope my ogas and generals haven't solve it sha...the answer to ur factorization is (x+2)(x+7)(5x+1)(4x-6) or we can say 2(x+2)(x+7)(5x+1)(x-3)....am coming back to show the workings.my b3 is down....#peace shalom#...happy maths solving.

@ smurfy please expand 2((2X-3)(5X+1)(X+7)(X+2))

smurfy: Again, you are wrong. The correct answer is 2(x + 7)(x + 2)(5x + 1)(2x - 3).

Now, what did you do wrong?

by simple algbraic expression the expression 20((x-3/2)(x+1/5)(x+7)(x+2)) can be express as 20((2x-3)(5x+1)(x+7)(x+2))/10 ; clearly we have 2((2x-3)(5x+1)(x+7)(x+2)

smurfy: Wrong.

@ smurfy the factorization is :20((x-3/2)(x+2)(x+7)(x+1/5)) *

benbuks: ..k here.

x is 1/3

smurfy: Factorise completely 20x^4 + 154x^3 + 40x^2 - 418x - 84.

20((x-3/2)(x+2)(x+7)(x+1/5))

benbuks: Try dis

@ ben buks hw did u post the question on paper please help me out

find the missing numbers : 2,_10,_40 :-(hint : _using the commutative law of addition

benbuks: ..i 4 fear na , i tought as much. He wanted to write log 3 _(x) + log x_(3) =10/3 nd nt 'log 3^x + log x^3 =10/3'

nice1 @FX...i respect o...

one can also apply change of base