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Nairaland Mathematics Clinic - Education (264) - Nairaland

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Re: Nairaland Mathematics Clinic by Nobody: 8:27am On Sep 13, 2019
MathsEconomics:


I have that book with me at home and I've read it before a good number of times.
I'll not lie to you, it's not a good book and it's best you stop wasting your time on it.
The book has little theoretical background but contains a lot of extremely difficult problems as exercises.
I suggest you stop using problem solving books for now. Use books like Chrystal's Algebra or Plane Trigonometry by Loney or Hobson (Both are equally good but Hobson is too rigorous for beginners).

The proofs presented in those Olympiad problem solving books will not really teach you how to construct your own proofs because they hide their steps and just give you the short and final answer leaving you with no hint as to how to construct your own proofs.

Also. As for the proofs you gave. I'll tell you something really important.

Just because you derived a valid inequality from a given inequality does not necessarily mean that that original inequality is valid.
For example. -4 < -3, -2 < -5 so by multiplication
8 < 15. Notice that the inequality we derived is valid but this is not a valid proof because our assertion that -2 < - 5 is false.
So regardless of how you manipulate inequalities, arriving at a true statement does not make that inequality true. That's why I dissected my proof above with induction to make it look like I'm actually deriving the result. Inequalities can be tricky.


Also. The conditions you imposed on you lemma actually invalidated it since the conditions you used do not satisfy the conditions imposed by the problem.
But if you're manipulating an inequality and the relationships between your steps(statements) are equivalences that means its allowed right?
Re: Nairaland Mathematics Clinic by Nobody: 9:07pm On Sep 13, 2019
Darivie04:

But if you're manipulating an inequality and the relationships between your steps(statements) are equivalences that means its allowed right?


Yes. So long as it's "equivalence" you mean and not "implication".
Re: Nairaland Mathematics Clinic by DrinkWater10: 9:06am On Sep 21, 2019
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Re: Nairaland Mathematics Clinic by Jackossky(m): 8:42pm On Sep 22, 2019
The question 3A please?

Re: Nairaland Mathematics Clinic by Brunicekid(m): 7:20am On Sep 23, 2019
Jackossky:
The question 3A please?
I will give you the solution TODAY by God's Grace.
Re: Nairaland Mathematics Clinic by Jackossky(m): 10:10am On Sep 23, 2019
Brunicekid:
I will give you the solution TODAY by God's Grace.

Will be expecting it...
Re: Nairaland Mathematics Clinic by Nobody: 10:37pm On Oct 07, 2019
Pls slv

Re: Nairaland Mathematics Clinic by Orestino(m): 9:14pm On Oct 10, 2019
Billyonaire369:
Pls slv
One of the solutions

Re: Nairaland Mathematics Clinic by Nobody: 11:47am On Oct 13, 2019
Richiez:
Hi everyone!
I'm Richard Obinna Agbara. We diagnose and solve math problems here.
This thread is the meeting point for Nairaland math gurus...I dare anyone to ask a question in mathematics without me having an answer to them, LETS START
pls solve for me

Re: Nairaland Mathematics Clinic by GMacbeth: 6:58am On Oct 23, 2019
Hep solve
Re: Nairaland Mathematics Clinic by naturalwaves: 3:40pm On Oct 23, 2019
Jackossky:
The question 3A please?
Let me get my pen and help you.
BRB

1 Like

Re: Nairaland Mathematics Clinic by Jackossky(m): 1:22pm On Oct 25, 2019
naturalwaves:

Let me get my pen and help you.
BRB

I think I figured it out. Would like to see yours to know of mine conflicts with yours.
Re: Nairaland Mathematics Clinic by GMacbeth: 4:47pm On Oct 25, 2019
GMacbeth:
Hep solve
Re: Nairaland Mathematics Clinic by dejt4u(m): 10:03pm On Oct 25, 2019
Jackossky:


I think I figured it out. Would like to see yours to know of mine conflicts with yours.
My answer is 45/101 = 0.4455
Re: Nairaland Mathematics Clinic by dejt4u(m): 10:17pm On Oct 25, 2019
Update:
Answer is (15/1000) × (30/100) = 45/10000 = 0.0045
Re: Nairaland Mathematics Clinic by Jackossky(m): 11:13pm On Oct 25, 2019
[quote author=dejt4u post=83464529]
My answer is 45/101 = 0.4455[/quote

AHH, please, I will like to see your workings.
Re: Nairaland Mathematics Clinic by bokunrawo(m): 7:22pm On Nov 06, 2019
Guys help

Re: Nairaland Mathematics Clinic by Mechanics96(m): 7:53pm On Nov 06, 2019
Monday
Re: Nairaland Mathematics Clinic by Mrshape: 8:37pm On Nov 06, 2019
bokunrawo:
Guys help
Yesterday=-1mod7
Today=0mod7
Tomorrow=1mod7
If -1mod7=1mod7 difference=2mod7
Then 0mod7=Saturday
Actual today=0mod7+2mod7=2mod7.
Actual today=Saturday +2days=Monday
Re: Nairaland Mathematics Clinic by Jackossky(m): 9:31am On Nov 08, 2019
dejt4u:

My answer is 45/101 = 0.4455


P(F/Cn) = 30%=0.3........... p(a)
P((F/Cm)= 70%=0.7.…...... P(b)
P(3/Cn) =15/1000=0.015. P(c)
P(3/Cm)=8/1000 =0.008. p(d)

From conditional probability P(F/G)= p(GnF)/p(F)


Then
the P( An b n C n D/C) = (0.3×0.7×0.015×0.008)/0.015
= 0.00168
=1.68/1000.

I'm kinda feeling that I've manipulated this result �
Re: Nairaland Mathematics Clinic by naturalwaves: 1:21pm On Nov 09, 2019
Jackossky:


I think I figured it out. Would like to see yours to know of mine conflicts with yours.
I am so sorry.
I have been just so so busy
Re: Nairaland Mathematics Clinic by tolulopevoice: 1:19pm On Nov 13, 2019
Mathematicians in the house solution to question 10 please

1 Share

Re: Nairaland Mathematics Clinic by Martinez39(m): 1:59pm On Nov 13, 2019
tolulopevoice:
Mathematicians in the house solution to question 10 please
The first important thing to observe is that the region described is one-eighth of the spherical region.

Given that R = 14 cm ,
1) V = (1/8 )(4/3)π(R³) = (11 × 392)/6 ≈ 718.67 cm³

2) The surface area of the region consists of one-eighth the surface of the sphere; two quarter circles of radii 14cm; and one-eighth of a circle of radius 14cm. (think deeply about this and calculate the area)
A = (7/8 )π(R²) = 539cm²

3) If 2k is for every cm², 539cm² will cost (2 × 539)k = 1078k


Cc. Mrshape, Darivie04
Re: Nairaland Mathematics Clinic by Mrshape: 3:15pm On Nov 13, 2019
Martinez39:

The first important thing to observe is that the region described is one-eighth of the spherical region.

Given that R = 14 cm ,
1) V = (1/8 )(4/3)π(R³) = (11 × 392)/6 ≈ 718.67 cm³

2) The surface area of the region consists of one-eighth the surface of the sphere; two quarter circles of radii 14cm; and one-eighth of a circle of radius 14cm. (think deeply about this and calculate the area)
A = (7/8 )π(R²) = 539cm²

3) If 2k is for every cm², 539cm² will cost (2 × 539)k = 1078k



Cc. Mrshape, Darivie04
I am in a class now I will check on it sonn
Re: Nairaland Mathematics Clinic by Nobody: 4:35pm On Nov 13, 2019
Martinez39:

The first important thing to observe is that the region described is one-eighth of the spherical region.

Given that R = 14 cm ,
1) V = (1/8 )(4/3)π(R³) = (11 × 392)/6 ≈ 718.67 cm³

2) The surface area of the region consists of one-eighth the surface of the sphere; two quarter circles of radii 14cm; and one-eighth of a circle of radius 14cm. (think deeply about this and calculate the area)
A = (7/8 )π(R²) = 539cm²

3) If 2k is for every cm², 539cm² will cost (2 × 539)k = 1078k


Cc. Mrshape, Darivie04
Lol I barely even understand the diagram
Re: Nairaland Mathematics Clinic by tolulopevoice: 4:42pm On Nov 13, 2019
Martinez39:

The first important thing to observe is that the region described is one-eighth of the spherical region.

Given that R = 14 cm ,
1) V = (1/8 )(4/3)π(R³) = (11 × 392)/6 ≈ 718.67 cm³

2) The surface area of the region consists of one-eighth the surface of the sphere; two quarter circles of radii 14cm; and one-eighth of a circle of radius 14cm. (think deeply about this and calculate the area)
A = (7/8 )π(R²) = 539cm²

3) If 2k is for every cm², 539cm² will cost (2 × 539)k = 1078k


Cc. Mrshape, Darivie04
thanks for the concern but the answer is wrong... For the volume they put 5030cm³, and for the surface area is 1925cm². And the cost 38.5 naira
Re: Nairaland Mathematics Clinic by Martinez39(m): 5:55pm On Nov 13, 2019
tolulopevoice:
thanks for the concern but the answer is wrong... For the volume they put 5030cm³, and for the surface area is 1925cm². And the cost 38.5 naira
Correction for me. My formula for volume is wrong and I miscalculated on that. The rest is correct. I await the input of other mathematicuans.

Given R = 14 cm² and π = 22/7

1) V = (1/8 )(2/3)π(R³) ≈ 718.67 cm³

2) A = (7/8 )π(R²) = 539 cm²

3) Cost of painting 539 cm² at 2k per cm² is (2 × 539)k = 1078k


Cc. Mrshape, darivie04
Re: Nairaland Mathematics Clinic by Mrshape: 6:33pm On Nov 13, 2019
Martinez39:
Correction for me. My formula for volume is wrong and I miscalculated on that. The rest is correct. I await the input of other mathematicuans.

Given R = 14 cm² and π = 22/7

1) V = (1/8 )(2/3)π(R³) ≈ 718.67 cm³

2) A = (7/8 )π(R²) = 539 cm²

3) Cost of painting 539 cm² at 2k per cm² is (2 × 539)k = 1078k


Cc. Mrshape, darivie04
The volume is 10800cm^3
Check again please,
Because you calculated for the volume of the
Cut out section only

1 Like 1 Share

Re: Nairaland Mathematics Clinic by Mrshape: 8:34pm On Nov 13, 2019
Martinez39:
Correction for me. My formula for volume is wrong and I miscalculated on that. The rest is correct. I await the input of other mathematicuans.

Given R = 14 cm² and π = 22/7

1) V = (1/8 )(2/3)π(R³) ≈ 718.67 cm³

2) A = (7/8 )π(R²) = 539 cm²

3) Cost of painting 539 cm² at 2k per cm² is (2 × 539)k = 1078k


Cc. Mrshape, darivie04

I am still at work if I get home,
I will bring my suggested solution with workings sir
Re: Nairaland Mathematics Clinic by Mrshape: 10:12pm On Nov 13, 2019
tolulopevoice:
Mathematicians in the house solution to question 10 please
The full shape is a share
Volume of a sphere is 4/3πr^3
Volume O SPHERE= 4/3×22/7×14^3
Volume of sphere=11494.04

But the portion cut out from it is 45° out of the hemisphere
So it will be 45/360 of hemisphere volume
Volume cross section =1/8×2/3×22/7×14^3
V CS=718.38

volume of solid=volume of sphere-volume of cross section
Volume of solid=11494.04-718.38
Volume of solid =10800cm^3
@martinez39
Re: Nairaland Mathematics Clinic by Mrshape: 10:18pm On Nov 13, 2019
TSA of solid =TSAsphere-TSA of outer cross section+ +TSA of inner criss section +area of roof and floor sector
TSA of solid=4πr^2-1/8×2πr^2+2/8 x2πr^2 + 2/8×πr^2
TSA of solid =4πr^2+3/8πr^2
=πr^2 (18/4)
=22/7×14^2×18/4
=2772
~2770cm^2
Re: Nairaland Mathematics Clinic by Mrshape: 10:31pm On Nov 13, 2019
At 2kobo per cm^2
Cost =(2/100)×2772
=#54:44

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