Joe82834:


The correct theorem states that,"If a secant and a tangent of a circle are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment. Interpreted as AC × BC = CT²
(6+x)6 = 8² = 64
6 + x = 64 ÷ 6 = 10.67
x = 10.67 - 6
x = 4.67

Check the bolded Bro....

Modified
Logically, considering the length of the line this seems more like it. seems like i omitted something in my approach

i mean d question u gave solutn to

Joe82834:


The correct theorem states that,"If a secant and a tangent of a circle are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment. Interpreted as AC × BC = CT²
(6+x)6 = 8² = 64
6 + x = 64 ÷ 6 = 10.67
x = 10.67 - 6
x = 4.67

Thank you very much. The little problem is that this theorem was not given in the text, so I believe there must be a simpler way of doing this using one of the eight theorems found in New General Maths for SS2.

Joe82834:


The correct theorem states that,"If a secant and a tangent of a circle are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment. Interpreted as AC × BC = CT²
(6+x)6 = 8² = 64
6 + x = 64 ÷ 6 = 10.67
x = 10.67 - 6
x = 4.67

Thank you very much. The little problem is that this theorem was not given in the text, so I believe there must be a simpler way of doing this using one of the eight ( found in New General Maths for SS2.

2nioshine:


Check the bolded Bro....

Modified
Logically, considering the length of the line this seems more like it. seems like i omitted something in my approach

Thank you very much. The little problem is that this theorem was not given in the text, so I believe there must be a simpler way of doing this using one of the eight found in New General Maths for SS2.

.

Cortana2:


Thank you very much. The little problem is that this theorem was not given in the text, so I believe there must be a simpler way of doing this using one of the eight ( found in New General Maths for SS2.

You can actually solve it without the theorem but it will take a longer process. join A and T with a line that passes through the centre (diameter), then join B and T as well and u get a chord. U will notice u get a right angle at B and a right angle at T from the known theorems. then u can get X by equating results got for AT from the two right-angled triangles formed. i.e CBT and ATB.

busuyem:


You can actually solve it without the theorem but it will take a longer process. join A and T with a line that passes through the centre (diameter), then join B and T as well and u get a chord. U will notice u get a right angle at B and a right angle at T from the known theorems. then u can get X by equating results got for AT from the two right-angled triangles formed. i.e CBT and ATB.

Thank you for replying late.
First, there is no way to convincingly prove that a line joining A and T passes through the centre of the circle. Secondly, I got a right angle at T but not at B, since B is not a point of tangency.

I would nevertheless appreciate if you can attach a picture of what you're trying to convey.

Thank you.

Cortana2:


Thank you for replying late.
First, there is no way to convincingly prove that a line joining A and T passes through the centre of the circle. Secondly, I got a right angle at T but not at B, since B is not a point of tangency.

I would nevertheless appreciate if you can attach a picture of what you're trying to convey.

Thank you.

Since A and T are on the circumference, that means u can draw two radii from them to meet at the centre which will form diametre. And if you get diameter from AT, that means angle ABT is right-angled and that settles it.

busuyem:


Since A and T are on the circumference, that means u can draw two radii from them to meet at the centre which will form diametre. And if you get diameter from AT, that means angle ABT is right-angled and that settles it.

A and B being on the circumference doesn't mean a straight line joining them must be a diameter,it could be a chord and that spoils d angle abt being a right angle theory.you are simple trying to use d eyes to assume dat line AT passes through d centre which isn't satisfactory.remember,every diameter is a chord but not every chord is a diameter

Joe82834:

A and B being on the circumference doesn't mean a straight line joining them must be a diameter,it could be a chord and that spoils d angle abt being a right angle theory.you are simple trying to use d eyes to assume dat line AT passes through d centre which isn't satisfactory.remember,every diameter is a chord but not every chord is a diameter

Ok. That was what I did then I got the answer. And I'm very sure that questn has solution without using that suggested theorem. Besides, if CT is tangent to the circle, it is certain that the tangent is perpendicular to the circle radius. If that is possible, can u extend the radius from T to A which must form a diameter since the extension passes through the center of the circle? Anyway, since u don't accept this, I think ur opinion counts. Bye.

busuyem:


Ok. That was what I did then I got the answer. And I'm very sure that questn has solution without using that suggested theorem. Besides, if CT is tangent to the circle, it is certain that the tangent is perpendicular to the circle radius. If that is possible, can u extend the radius from T to A which must form a diameter since the extension passes through the center of the circle? Anyway, since u don't accept this, I think ur opinion counts. Bye.

Hope u no de vex,cos ppl argue to learn.
Definitely d tangent is perpendicular to d radius but nothing (except with d eyes) tells us that a straight line from d tangent to d centre, if extended passes through A.

Joe82834:

Hope u no de vex,cos ppl argue to learn.
Definitely d tangent is perpendicular to d radius but nothing (except with d eyes) tells us that a straight line from d tangent to d centre, if extended passes through A.

Lolz...I no dey vex but I no fit write online like that. Are u trying to say u can't get a triangle through points ATB and if u can, do u mean it can't be possible for T to join A through the centre? Better still, have u used what I ealier said to solve the question and get it? It's like u didn't bother to use at all because if u did, u would get better explanation urself than written here. I only responded to your questn because u mentioned General Maths as the texbk u used which I also used while teaching Maths in senior classes. Anyway, I might not be right afterall because this is Maths which has no mr know-it-all. Even the theorem used to solve that questn came out of this because it's never one of the fundamental theorems of cirgle theorms.
#edited

Please, I need someone to solve this maths:

1. A store determines the retail price of a DVD Player by marking up its wholesale price by 60 percent. After a 25 percent discount off the DVD Player's retail price, the DVD Player cost $162. What is the wholesale price of the DVD Player?

2. Evaluate 6P3-6C3

Mrphantom99:
Please, I need someone to solve this maths:

1. A store determines the retail price of a DVD Player by marking up its wholesale price by 60 percent. After a 25 percent discount off the DVD Player's retail price, the DVD Player cost $162. What is the wholesale price of the DVD Player?

2. Evaluate 6P3-6C3

.1 #135

I love dis Forum, pls direct some question to me, if u have any problem with math.

Dis is my watapp no 08133550169

Mrphantom99:
Please, I need someone to solve this maths:

1. A store determines the retail price of a DVD Player by marking up its wholesale price by 60 percent. After a 25 percent discount off the DVD Player's retail price, the DVD Player cost $162. What is the wholesale price of the DVD Player?

2. Evaluate 6P3-6C3

1. (x + 60% of x) - 25% of x = 162
x + 3/5x - 1/4x = 162
1.35x = 162
x = 162/1.35
x = N120

Wholesale price of the DVD is N120

2. 6P3 = 6! / (6-3)!
6! / 3!
6*5*4 = 120

6C3 = 6!/3!3!
6*5*4 / 3*2*1 = 20

6P3-6C3
120 - 20 = 100

Mrphantom99:
Please, I need someone to solve this maths:

1. A store determines the retail price of a DVD Player by marking up its wholesale price by 60 percent. After a 25 percent discount off the DVD Player's retail price, the DVD Player cost $162. What is the wholesale price of the DVD Player?

2. Evaluate 6P3-6C3

Mathemagician1 has solved question 1. However your question 2 isn't clear. Do you mean simplify or evaluate?

Obinoscopy:

Mathemagician1 has solved question 1. However your question 2 isn't clear. Do you mean simplify or evaluate?

I've solved both. I was working on the second question before your msg. Cheers

Mathemagician1:


[s]1. (x + 60% of x) - 25% of x = 162
x + 3/5x - 1/4x = 162
1.35x = 162
x = 162/1.35
x = N120

Wholesale price of the DVD is N120[/s]

2. 6P3 = 6! / (6-3)!
6! / 3!
6*5*4 = 120

6C3 = 6!/3!3!
6*5*4 / 3*2*1 = 20

6P3-6C3
120 - 20 = 100

1. (x+0.6x)*0.75=162
1.6x*0.75=162
1.2x=162
x=162/1.2
x=135
@ fawaz050 is right

ladokuntlad:

1. (x+0.6x)*0.75=162 1.6x*0.75=162 1.2x=162 x=162/1.2 x=135 @ fawaz050 is right

boss longtym fa..u just ignore ur students

fawaz050:
boss longtym fa..u just ignore ur students

Life for here dey roughly smooth oo
I no ignore una ooo.

U know there is real life after school which is what man is facing now

ladokuntlad:

Life for here dey roughly smooth oo I no ignore una ooo.
U know there is real life after school which is what man is facing now

E go work out sir

Richiez:
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

An integer P is a ............iff
(p-1)! is congruence to -1(modp) fill in the missing word and provide the name of the Mathematician that gave that definition........ Happy Sunday to you.

Agnesgrace:

An integer P is a ............iff
(p-1)! is congruence to -1(modp) fill in the missing word and provide the name of the Mathematician that gave that definition........ Happy Sunday to you.

congruence modulo by Euler

ladokuntlad:


congruence modulo by Euler

No. Euler definition for congruence is different from that........ Good morning and have a wonderful day.

busuyem:


Since A and T are on the circumference, that means u can draw two radii from them to meet at the centre which will form diametre. And if you get diameter from AT, that means angle ABT is right-angled and that settles it.

The one at T will be form a right angle with the radius but the one at A will not. Remember the theorem says: A tangent to a circle is perpendicular to the radius drawn to its point.

Long time here

I love this thread,I have been searching for this kind of thing on facebook,the one I came in contact with they just tell u the answer,so glad I found this thread.

Gurus in the house, please help me with diz question on sum of geometric progression... IN AN EXPONENTIAL SERIES OF INCREASING POSITIVE TERMS, THE SUM OF THE FIRST SIX TERMS IS NINE TIMES THE SUM OF THE FIRST THREE TERMS. THE SUM OF THE FOURTH AND THE FIFTH TERM IS 120. FIND THE
(1) CONSTANT RATIO
(2) FIRST TERM
(3) SUM OF THE FIRST TWENTY TERMS.

Hadampson:
[i][/i] gurus in the house, abeg help me with diz question on sum of geometric progression... IN AN EXPONENTIAL SERIES OF INCREASING POSITIVE TERMS, THE SUM OF THE FIRST SIX TERMS IS NINE TIMES THE SUM OF THE FIRST THREE TERMS. THE SUM OF THE FOURTH AND THE FIFTH TERM IS 120. FIND THE (1) CONSTANT RATIO (2) FIRST TERM (3) SUM OF THE FIRST TWENTY TERMS.

.