donedy:

SOLUTIONS

From equations of motion: S = V_it + 1/2*a*t² ----- Eq.1

a = -9.81 m/s {gravitational acceleration}
V_i = Given {Initial velocity}
t = 30 s { total elapse time)
S = ? { Change in Distance covered within the given time)

---
^
|
| H
|
|
|
|
------------------------ Ground

a) From Eq.1, V_i = 0 since the balloon was at rest when the bomb was dropped. So,

S = 0 *(30 s) + 1/2 *(-9.81 m/s)*(30 s)² = - 4414.5 m

H = -S

Thus height, H = 4414.5 m

b) From Eq.1, V_i = 100 cm/s = 1 m/s since the balloon was ascending at that speed.

S = (1 m/s)*(30 s) + 1/2 *(-9.81 m/s)*(30 s)² = -4384.5 m

H = -S

Thus height, H = 4384.5 m

Wow! Thanks a lot bro,I owe yu 1

Richiez: @WHEELS, thanks for your compliments. my gurus will devour your questions very soon, but i'd advise u post them here for speedy answer

I thought about that earlier but i think one can understand the question more if he/she thoroughly examine the nature of the question by visiting the sister-thread . Anyway i will post the question as you suggested and if one needs further digging he shall just view the sister-threads content and how other people try to deal with the question b4.

It goest this way:
3x+2y-7=19 ------(1)
4x-y+27=4 --------(2)
2x+4y-52=32 -------(3)

in this case collecting like terms wont work cos the author of this question want us to find the unknown variables preferably using the elimination method.. I know it looks peculiar[cos 3 variables need to be there not 2] but the author manage to provide solution out of this obfuscating equation.. Hence you knw of the question, now i will stick 2 dis thread for a while and see how our maths gurus will get d numbers of the unknown[3] variables[z included]..

Check the siser thread here: www.nairaland.com/1199966/maths-homework-please

WHξξ∟s:


I thought about that earlier but i think one can understand the question more if he/she thoroughly examine the nature of the question by visiting the sister-thread . Anyway i will post the question as you suggested and if one needs further digging he shall just view the sister-threads content and how other people try to deal with the question b4.

It goest this way:
3x+2y-7=19 ------(1)
4x-y+27=4 --------(2)
2x+4y-52=32 -------(3)

in this case collecting like terms wont work cos the author of this question want us to find the unknown variables preferably using the elimination method.. I know it looks peculiar[cos 3 variables need to be there not 2] but the author manage to provide solution out of this obfuscating equation.. Hence you knw of the question, now i will stick 2 dis thread for a while and see how our maths gurus will get d numbers of the unknown[3] variables[z included]..

Check the siser thread here: www.nairaland.com/1199966/maths-homework-please

Hey, what's this?

@richiez, u still havnt postd d mathematical induction soln

ositadima1:

Hey, what's this?

LOL! Am glad that other people neva come across a thing like this b4, i will now confidently conclude that it is a typographical error.. Anyway, what do u tink about the equation?

yungryce: @richiez, u still havnt postd d mathematical induction soln

oh i'm rily rily sorry pls repost the question

WHξξ∟s:
LOL! Am glad that other people neva come across a thing like this b4, i will now confidently conclude that it is a typographical error.. Anyway, what do u tink about the equation?

first, only 2 variables can be seen.
secondly, only 2 equations is needed for two unknowns, meaning one of the equations is irrelevant

Na im b dis
Prove by induction that for any two nos a & b, & for any nEN that
(a+b)^n = a^n + na^(n-1)b + [n(n-1)/2!]a^(n-2)b^(2) + ...+ [n!/(n-k)!k!]a^(n-k)b^k + ... + b^(k)

yungryce: Na im b dis Prove by induction that for any two nos a & b, & for any nEN that (a+b)^n = a^n + na^(n-1)b + [n(n-1)/2!]a^(n-2)b^(2) + ...+ [n!/(n-k)!k!]a^(n-k)b^k + ... + b^(k)

check the solution below

WHξξ∟s:


LOL! Am glad that other people neva come across a thing like this b4, i will now confidently conclude that it is a typographical error.. Anyway, what do u tink about the equation?

A lot of errors in the question! Logically wrong....

Prove by induction that for any two nos a & b, & for any nEN that
(a+b)^n = a^n + na^(n-1)b + [n(n-1)/2!]a^(n-2)b^(2) + ...+ [n!/(n-k)!k!]a^(n-k)b^k + ... + b^(k)

SOLUTION

first, let us show that the statement is true for n=1


(a+b)^1 = a^1 + 1[a^(1-1)b] + 0 + 0 +....


a+b = a+b.....true


Now we can assume that it is true for n=k


(a+b)^k = a^k + ka^(k-1)b + k(k-1)/2![a^(k-2)]b^2 +...+...


similarly, For n=k+1


(a+b)^(k+1) = a^(k+1) +(k+1)a^(k+1-1)b +(k+1)(k+1-1)/2![a^(k+1-2)b^2]+.....


(a+b)^(k+1) = a^(k+1) + (k+1)a^k(b) + k(k+1)/2![a^(k-1)b^2]+.....


To verify this, multiply both sides of n=k by (a+b) to get same expression as in n=k+1


(a+b)^k.(a+b) = a^k.(a+b) + ka^(k-1)b.(a+b) +k(k-1)/2


(a+b)^(k+1) = a.a^k + b.a^k + a.k.a^(k-1)b + b.k.a^(k-1)b + a.k(k-1)/2![a^(k-2)b^2] + b.k(k-1)/2![a^(k-2)b^2]


(a+b)^(k+1) = a^(k+1) + b.a^k + ka^k.b +ka^(k-1)b^2 + k(k-1)/2![a^(k-1)b^2] + k(k-1)/2![a^(k-2)b^3]


The b^3 term will be added to the next term in the series Therefore;

(a+b)^(k+1) = a^(k+1) + (k+1)a^k(b) + [k+ k(k-1)/2!]a^(k-1)b^2

by factorizing,

(a+b)^(k+1) = a^(k+1) + (k+1)a^k(b) + k(k+1)/2![a^(k-1)b^2]+......

Now, since this last equation is same for n=k+1 , we have proven the validity of the statement by induction.

@youngryce note that for purpose of clarity, the b^3 term was ignored because it will be added to the next term in the series with 'b^3' hence the use of +.....

Goodday all the gurus here, please i have a little problem:
solve 2^X=30. Find X?
Solution.
Taking log to base 2 of both sides
logbase2 (2^X)=logbase2 (30)
Xlogbase2 (2)=logbase2 (30)
and a law states that
Xlogbase p (p)=X*1
therefore:
X*1=logbase2 (30)
X=logbase2 (30)
please where do i go from here?

johnpaul1101: Goodday all the gurus here, please i have a little problem: solve 2^X=30. Find X? Solution. Taking log to base 2 of both sides logbase2 (2^X)=logbase2 (30) Xlogbase2 (2)=logbase2 (30) and a law states that Xlogbase p (p)=X*1 therefore: X*1=logbase2 (30) X=logbase2 (30) please where do i go from here?

Good work Johnpaul. well since it is easier to evaluate in base10 [because our calculators and four figure tables are readily available in base10]
we'll use another law

Logbase(a)(b) = Logbase(x)b/Logbase(x)a
therefore

Logbase2(30) = Logbase10(30)/Logbase10(2)

= 1.477/0.3010 = 4.907

Richiez:

Good work Johnpaul. well since it is easier to evaluate in base10 [because our calculators and four figure tables are readily available in base10]
we'll use another law

Logbase[b]a[/b]b = Logbase(x)b/Logbase(x)a
therefore
Logbase2(30) = Logbase10(30)/Logbase10(2) =

please dont be offended @richiez, i dont understand the law you just used.

johnpaul1101: please dont be offended @richiez, i dont understand the law you just used.

This law allows us to change to a suitable base.
notice how we choose a suitable base (base10).
now the numerator will have the original number, while the denominator the former base.
e.g we all know that Logbase(2)4 = 2.
Now this is same as Logbase(10)4/Logbase(10)2 = 2
try using a scientific calculator to check this out

Plz...guyz check dis out...find d general solutions of (1) 7sec^2€= 6tan€ + 8...(2) 2cos^2€-3sin2€-2...where ^ and € means power and tetha respectively..i.e 2cos squre tetha= 2cos^2€...tanx alot

Fetus: Plz...guyz check dis out...find d general solutions of (1) 7sec^2€= 6tan€ + 8...(2) 2cos^2€-3sin2€-2...where ^ and € means power and tetha respectively..i.e 2cos squre tetha= 2cos^2€...tanx alot

SOLUTION
QUESTION 1
7sec^2€ = 6tan€ + 8 7sec^2€ -6tan€ -8 = 0 ..........(1)
recall that sec^2€ = tan^2€ + 1, now puttin this in eqn(1) gives;
7(tan^2€ + 1) -6tan€ -8 =0
7tan^2€ + 7 -6tan€ -8 =0
7tan^2€ -6tan€ -1 =0 let tan€ = y
7y^2 -6y -1 =0 .........(2)
eqn(2) is now a quadratic eqn which can be solved easily. 7y^2 -7y +y -1 =0 7y(y-1) +1(y-1) =0
(y-1)(7y+1) =0
y= 1 or -1/7
recall the substitution tan€=y
therefore;
tan€ = 1 or -1/7 when tan€=1
€ = tan^-11 €= 45^¤ or 225^¤ when tan€= -1/7
€= tan^-1-1/7
€ = 171.89^¤ or 351.87^¤

@fetus check question2, i think the 'equal to' sign was omitted. however u can now solve it yourself using question1 as a guide
use this hint; cos^2€ = 1-sin^2€

Richiez:
This law allows us to change to a suitable base.
notice how we choose a suitable base (base10).
now the numerator will have the original number, while the denominator the former base.
e.g we all know that Logbase(2)4 = 2.
Now this is same as Logbase(10)4/Logbase(10)2 = 2
try using a scientific calculator to check this out

Thanks a lot richiez, it helped

good job guys keeping the thread!

biolabee: good job guys keeping the thread!

Na your eye be this?

please help me with this:
if cos2§=sin3§, find §?
(note-§ represents teta)

biolabee: good job guys keeping the thread!

where av u been?

johnpaul1101: please help me with this:
if cos2§=sin3§, find §?
(note-§ represents teta)

soln
cos2@ = sin3@
cos2@ can also be written as sin (90-2@)
this implies dat sin (90-2@) = sin3@
therefore, 90-2@=3@
90=3@+2@
90=5@
therefore, @=18 degree. Since we are not asked to find the values (i.e. not one value) of @, the general formula may not be applicable. Note: '@' is used to rep. theta

@bolaji, nice work, welcome to the thread.

Bolaji 16: soln
cos2@ = sin3@
cos2@ can also be written as sin (90-2@)
this implies dat sin (90-2@) = sin3@
therefore, 90-2@=3@
90=3@+2@
90=5@
therefore, @=18 degree. Since we are not asked to find the values (i.e. not one value) of @, the general formula may not be applicable. Note: '@' is used to rep. theta

thanks a lot

where my gurus at?
If Y=tan^-1x, find dy/dx

Richiez: where my gurus at?
If Y=tan^-1x, find dy/dx

solution
y = tan-1x
tan y = x
sec2y*dy/dx = 1
dy/dx = 1/sec2y
=1/1+tan2y
=1/1+(tany)^2
=1/1+x^2

Hello maths guru, pls I need solution to this question. Thanks in advance.


A plane files northwards for 430km, it then flies eastwards for 380km. How far is it from the starting point?

jhydebaba: Hello maths guru, pls I need solution to this question. Thanks in advance.


A plane files northwards for 430km, it then flies eastwards for 380km. How far is it from the starting point?

i believe that i am still a learner, i still think i'm entitled to trial, so if i may contribute, i would say:
since this is bearing, after interpreting the question, i arrived at 574km(using pythagoras theorem). Please gurus, am i wrong?

johnpaul1101:
i believe that i am still a learner, i still think i'm entitled to trial, so if i may contribute, i would say:
since this is bearing, after interpreting the question, i arrived at 574km(using pythagoras theorem). Please gurus, am i wrong?

ok