₦airaland Forum

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.

Darivie04:
But if you're manipulating an inequality and the relationships between your steps(statements) are equivalences that means its allowed right?

Jackossky:
The question 3A please?

Brunicekid:
I will give you the solution TODAY by God's Grace.

Billyonaire369:
Pls slv

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

Jackossky:
The question 3A please?

naturalwaves:
Let me get my pen and help you.
BRB

GMacbeth:
Hep solve

Jackossky:
I think I figured it out. Would like to see yours to know of mine conflicts with yours.

bokunrawo:
Guys help

dejt4u:
My answer is 45/101 = 0.4455

Jackossky:
I think I figured it out. Would like to see yours to know of mine conflicts with yours.

tolulopevoice:
Mathematicians in the house solution to question 10 please

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

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

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

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

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

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

tolulopevoice:
Mathematicians in the house solution to question 10 please