dprinz99:
Pls solve

1)1/8.7^-3=y^3
find y

2)(root5-2)^2-z=(1-root20)^2.
find z

Correct the questions if I'm mistaken.

1.

1/8 * 7^-3 = y³

2^-3 * 7^-3 = y³

(2*7)^-3 = (1/y)^-3

2*7 = 1/y

y = 1/14


2.


(√5 -2)² - z = (1 - √20)²

z = (√5 -2)² - (1 - √20)²

z = 5 - 4√5 + 4 - 1 + 2√20 - 20

z = -12 - 4√5 + 2√20

z = -12 - 4√5 + 4√5

z = -12

somebody in the house should please help me expand (x + 4)^3 and (x + 2)^4 Bionomially please guys its urgent. thanks

youngdude1:
somebody in the house should please help me expand (x + 4)^3 and (x + 2)^4 Bionically please guys its urgent. thanks

Bionically??
I guess you mean Binomially? Lol

DatechMan:


Bionically??
I guess you mean Binomially? Lol

YES..that was a typo *modified*

youngdude1:
somebody in the house should please help me expand (x + 4)^3 and (x + 2)^4 Bionomially please guys its urgent. thanks

use Pascal's triangle or Combination method.
Hint: for power 3-----1-3-3-1(Pascal''s triangle method). Opps, low battery.

thankyouJesus:
use Pascal's triangle or Combination method.
Hint: for power 3-----1-3-3-1(Pascal''s triangle method). Opps, low battery.

i don't get you

youngdude1:
somebody in the house should please help me expand (x + 4)^3 and (x + 2)^4 Bionomially please guys its urgent. thanks

(x + 4)³

x³ + ³C₁ x².4 + ³C₂ x.4² +3³

x³ + 12x² + 48x + 27

DatechMan:




It is not an assumption. Check the solution again.

Note:

0.1< k ≤ 1

Therefore -1 ≤ log k ≤ 0

This implies that we will always round up

Thıs observatıon ıs especıally for u is prime where u is the power of exponetial expression for which the total number of digits are to be found, try 3^129, 3^17 though logk<1 but there are cases where logk is very close to 1, thus altering the actual approximation of the number of digits

DatechMan:



(x + 4)³

x³ + ³C₁ x².4 + ³C₂ x.4² +3³

x³ + 12x² + 48x + 27

is that C or a bracket? Thanks

youngdude1:
is that C or a bracket? Thanks

C is Combination.

benji93:

Thıs observatıon ıs especıally for u is prime where u is the power of exponetial expression for which the total number of digits are to be found, try 3^129, 3^17 though logk<1 but there are cases where logk is very close to 1, thus altering the actual approximation of the number of digits

Yes, I must admit there are cases where it is extremely close. But I don't think there exist a situation where it is actually 1.

There's an exception though. When the index is 1. e.g 1²⁵⁰

For the instances above:

3¹²⁹ has 62 digits.

129log3 = 61.5486... (Rounded up, we have 62) valid

3¹⁷ has 9 digits.

17log3 = 8.111... (Rounded up, we have 9) valid



Good Morning Nigeria

#GodBlessNigeria

ElDeeVee:

C is Combination.

thanks

DatechMan:

Yes, I must admit there are cases where it is extremely close. But I don't think there exist a situation where it is actually 1.

There's an exception though. When the index is 1. e.g 1²⁵⁰

For the instances above:

3¹²⁹ has 62 digits.

129log3 = 61.5486... (Rounded up, we have 62) valid

3¹⁷ has 9 digits.

17log3 = 8.111... (Rounded up, we have 9) valid



Good Morning Nigeria

#GodBlessNigeria

Oh isee thts ur idea of rounding up
and ofcourse not there is never a case where it is 0.1 or 1 exactly
it comes from the idea that it comes from the fact that every exponential expression a^b>1 is such that 10^n-1<a^b>10^n infact it was from this axiom i thought that a^b could be at any point at infinity very close to the lower boundary or the upper boundary, if it is very close to the upper boundary, where logk <1 normally based on your rounding up if you had n=999.62 forexample it would be 1000 but actually if the value of k is to considered you could have 999.62 + 0.4 which gives 1000.02 which should be approximated to 1001 somehow we would not be able to test all numbers to infinity, but we can test as many number of digits as possible, i am not trying to disprove you bro,i am just saying based on my assumption there is a possibilty of having that +-1 problem somewhere at infinity unless my assumption is wrong which is also another possibility, we are in a mathematical world, there are just too many things that are not exactly predictable.All the best

Youngsage:

x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
log(x+y) = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.

. Please is logx + log y the same as log(x+y) recalling that log2 + log 3 is equal to log(2 x 3) which is not equal to log(2+3)

Wonderful forum...Happy new year to you all.

Agnesgrace:
Wonderful forum...Happy new year to you all.

Same to you dear.

About your post about Logx + logy na mistake. The guy dey dream.

Welcome on board


#GodBlessNigeria

goodevening all, plz i need ya help here: diffferentiate d fllwn;
(1).....Y=sqre root(X+X^2)
(2)....Y=(3X^4)-(X^-1).

Please somebody should help me with this question it is urgent pls.
The third term of a geometric progressiom is nine
times the first term.if the second term is one-twenty
fourth the fifth term.find the fourth term.ar^2=9a.
r=sqr of 9 r=3.pls teach me how to get the first term

This thread has made some people to be lazy. People come here to ask questions without extra efforts to get them solved.

I think before any question(s) is been solved on this thread. You will need to upload your solved solution and people on this thread will then guide you to solve them.

johnpaul1101:
goodevening all, plz i need ya help here: diffferentiate d fllwn;
(1).....Y=sqre root(X+X^2)

y= (x+x^2)^0.5

let u= x+x^2
=> y= u^0.5
dy/du = 0.5u^-0.5 = 1/(2√u)

du/dx = 1+2x

thus , dy/dx = du/dx * dy/du

=> dy/dx = (1+2x)* 1/(2√u)

hence dy/dx= (1+2x)/[2√(x+x^2) ]

(2)....Y=(3X^4)-(X^-1).

dy/dx = 12x^3 + x^-2


guess that's it.....hope u get
[/quote]

Ajibade105:
This thread has made some people to be lazy. People come here to ask questions without extra efforts to get them solved.

I think before any question(s) is been solved on this thread. You will need to upload your solved solution and people on this thread will then guide you to solve them.

exactly the point sir.. .I just taya..4 dem....

its well sha.

Agnesgrace:
Please somebody should help me with this question it is urgent pls.
The third term of a geometric progressiom is nine
times the first term.if the second term is one-twenty
fourth the fifth term.find the fourth term.ar^2=9a.
r=sqr of 9 r=3.pls teach me how to get the first term

. Pls much effort hd been made to solve this problem. Only r is obtainable pls if u know the first term help pls

agentofchange1:

(2)....Y=(3X^4)-(X^-1).
dy/dx = 12x^3 + x^-2

guess that's it.....hope u get

I was actually hoping u'd do dat using the 2nd-p.

Agnesgrace:
. Pls much effort hd been made to solve this problem. Only r is obtainable pls if u know the first term help pls

Find a paper write the question on it, snap n upload it here

Agnesgrace:
. Pls much effort hd been made to solve this problem. Only r is obtainable pls if u know the first term help pls

You question is seemingly inconsistent.

For a G.P the ratio of T₅ to T₂ is always r³.

Therefore I can't figure out why T₂ is ¹/₂₄ of T₅.

Kindly double check the question. Is there something I'm missing?

Also note that knowing the ratio of terms in GP cannot give us the value of the first term, I think more information are needed

DatechMan:


You question is seemingly inconsistent.

For a G.P the ratio of T₅ to T₂ is always r³.

Therefore I can't figure out why T₂ is ¹/₂₄ of T₅.

Kindly double check the question. Is there something I'm missing?

Also note that knowing the ratio of terms in GP cannot give us the value of the first term, I think more information are needed

. Thanks very much, that is how the question ís hv checked again and again pl i dont know if there is any other approach

johnpaul1101:
goodevening all, plz i need ya help here: diffferentiate d fllwn;
(1).....Y=sqre root(X+X^2)
(2)....Y=(3X^4)-(X^-1).

FYI.
I had to jump some steps. Hope the algorithm is clear

Agnesgrace:
. Thanks very much, that is how the question ís hv checked again and again pl i dont know if there is any other approach

Thanks very much great people i hd done justice to that. Have a wonderful night rest. Bye

Agnesgrace:
Thanks very much great people i hd done justice to that. Have a wonderful night rest. Bye

Would you be kind enough to share with us?

johnpaul1101:
I was actually hoping u'd do dat using the 2nd-p.

you didn't indicate what's to be used
BTW: what's. '2nd-p' ?

agentofchange1:


you didn't indicate what's to be used

BTW: what's. '2nd-p' ?

Maybe 2nd-p = Second Principle

Please forgive my ignorance. johnpaul1101 kindly shed light

DatechMan:

Maybe 2nd-p = Second Principle
Please forgive my ignorance. johnpaul1101 kindly shed light

lolz. ..haha.. I wanted to say so......just wanted to be sure... ....
we learn everyday u kw...

#Happy.Sunday.