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Re: Nairaland Mathematics Clinic by tolulopevoice: 5:28pm On Nov 14, 2019
Mathematicians please solution to question 5
Re: Nairaland Mathematics Clinic by Martinez39(m): 5:54pm On Nov 14, 2019
tolulopevoice:
Mathematicians please solution to question 5

Use the special formula for the difference of two squares.
P² - Q² = (P - Q)(P + Q)

P½ - Q½ = (P¼ - Q¼)(P¼ + Q¼)
Given that
M½ = √(M)

(P½ - Q½)(√(P) +√(Q) ) = (P½ - Q½)(P½ + Q½) = P - Q
Re: Nairaland Mathematics Clinic by dejt4u(m): 5:55pm On Nov 14, 2019
tolulopevoice:
Mathematicians please solution to question 5

(p¼ - q¼)(p¼ + q¼)(√p + √q)
= (p¼ - q¼)(p¼ + q¼)(p½ + q½)
= {(p¼)(p¼) + (p¼)(q¼) - (p¼)(q¼) - (q¼)(q¼)}{p½ + q½}
= {(p¼)(p¼) - (q¼)(q¼)}{p½ + q½}
= {p½ - q½}{p½ + q½}
= (p½)(p½) + (p½)(q½) - (p½)(q½) - (q½)(q½)
= (p½)(p½) - (q½)(q½)
= p - q

2 Likes

Re: Nairaland Mathematics Clinic by tolulopevoice: 10:05pm On Nov 14, 2019
dejt4u:

(p¼ - q¼)(p¼ + q¼)(√p + √q)
= (p¼ - q¼)(p¼ + q¼)(p½ + q½)
= {(p¼)(p¼) + (p¼)(q¼) - (p¼)(q¼) - (q¼)(q¼)}{p½ + q½}
= {(p¼)(p¼) - (q¼)(q¼)}{p½ + q½}
= {p½ - q½}{p½ + q½}
= (p½)(p½) + (p½)(q½) - (p½)(q½) - (q½)(q½)
= (p½)(p½) - (q½)(q½)
= p - q
thanks
Re: Nairaland Mathematics Clinic by tolulopevoice: 10:06pm On Nov 14, 2019
tolulopevoice:
Mathematicians in the house solution to question 10 please
still waiting for correct solution to this please
Re: Nairaland Mathematics Clinic by Mrshape: 10:22pm On Nov 14, 2019
tolulopevoice:
still waiting for correct solution to this please
Tell me which of my solution is not correct?
Re: Nairaland Mathematics Clinic by Mrshape: 10:27pm On Nov 14, 2019
tolulopevoice:
still waiting for correct solution to this please
Tell me which of my solution is not correct?
Let me look at the question again
Re: Nairaland Mathematics Clinic by tolulopevoice: 10:30pm On Nov 14, 2019
Mrshape:

Tell me which of my solution is not correct?
Let me look at the question again

Re: Nairaland Mathematics Clinic by Mrshape: 10:36pm On Nov 14, 2019
[quote author=tolulopevoice post=84035124][/quote]
Check my previous solution I have attempted it
Re: Nairaland Mathematics Clinic by tolulopevoice: 9:22pm On Nov 15, 2019
Mrshape:

Check my previous solution I have attempted it
the answer at the textbook says the volume is 5030cm³
Re: Nairaland Mathematics Clinic by Mrshape: 9:31pm On Nov 15, 2019
tolulopevoice:
the answer at the textbook says the volume is 5030cm³
I know what they did,
The assume that same is an hemisphere, although they told us but we can see a sphere their
Re: Nairaland Mathematics Clinic by Mrshape: 9:33pm On Nov 15, 2019
tolulopevoice:
the answer at the textbook says the volume is 5030cm³
7/8 πr^3=7/8×22/7×14^3=5028.84~5030cm^3
Re: Nairaland Mathematics Clinic by Mrshape: 9:46pm On Nov 15, 2019
tolulopevoice:
the answer at the textbook says the volume is 5030cm³
TSA of solid =TSAhemisphere-TSA of outer cross section+ area of two quadrants +area of roof sector

TSA of solid=2πr^2-1/8×2πr^2+1/2πr^2 + 7/8×πr^2

TSA of solid =3.125πr^2

=26900cm^2
Re: Nairaland Mathematics Clinic by tolulopevoice: 10:46pm On Nov 15, 2019
Mrshape:

7/8 πr^3=7/8×22/7×14^3=5028.84~5030cm^3
please can you help with proper and clear solutions
Re: Nairaland Mathematics Clinic by Mrshape: 10:50pm On Nov 15, 2019
tolulopevoice:
please can you help with proper and clear solutions
Your problem is that you don't respond quick
I am solving this thing using mental work, so their is high chance of error,
If you respond quick I will know if you check my solution again,
But your response is every 12 hours.
Re: Nairaland Mathematics Clinic by Mrshape: 10:56pm On Nov 15, 2019
tolulopevoice:
please can you help with proper and clear solutions
full shape is a share

Volume of a hemisphere is 2/3πr^3

Volume 2/3×22/7×14^3

Volume =5747cm^3



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 hemisphere-volume of cross section

Volume of solid=5747-718.38

Volume of solid =5030cm^3
Re: Nairaland Mathematics Clinic by Akanoo1: 6:00pm On Nov 18, 2019
In how many ways can the letters of the word OBASANJO be arranged such that the vowels will not be together. Need help on this question please, I promised my cousin to assist. Thank you
Re: Nairaland Mathematics Clinic by Mrshape: 6:39pm On Nov 18, 2019
Akanoo1:
In how many ways can the letters of the word OBASANJO be arranged such that the vowels will not be together. Need help on this question please, I promised my cousin to assist. Thank you
Lol promise and fail.
Add one million dollars to it and see how we go rush am
Re: Nairaland Mathematics Clinic by Akanoo1: 7:20pm On Nov 18, 2019
Mrshape:

Lol promise and fail.
Add one million dollars to it and see how we go rush am
I know there's a platform like this before I said that. Kindly help.
Re: Nairaland Mathematics Clinic by naturalwaves: 9:53pm On Nov 18, 2019
Akanoo1:
In how many ways can the letters of the word OBASANJO be arranged such that the vowels will not be together. Need help on this question please, I promised my cousin to assist. Thank you

Step 1.

You have to find the total number of ways the entire words can be arranged and that is 8!/(2! 2!)- This is because letters O and A are doubled.

8!/(2!2!)= 1680 ways


Step 2

You have to find the number of ways the word can be arranged if the vowels must be together. E. G
(OAAO) BSNJ.

This implies that the vowels are 1 letter and we have 4 other letters which are BSNJ

Number of arrangement in this scenerio is 5!= 120

But we also know that each of the vowels can swap positions in this arrangement. Each one can maintain 4 different positions. Thus, the the number of ways of arrangement if the vowels must be together is 120 x 4 =480 ways

Finally, the number of ways of arranging the name OBASANJO if the vowels must not be together = Total number of arrangements possible - Total number of arrangements when vowels are together

This implies 1680-480
Therefore, answer = 1200 ways

I hope this helps?

4 Likes

Re: Nairaland Mathematics Clinic by Mrshape: 10:03pm On Nov 18, 2019
naturalwaves:


Step 1.

You have to find the total number of ways the entire words can be arranged and that is 8!/(2! 2!)- This is because letters O and A are doubled.

8!/(2!2!)= 1680 ways


Step 2

You have to find the number of ways the word can be arranged if the vowels must be together. E. G
(OAAO) BSNJ.

This implies that the vowels are 1 letter and we have 4 other letters which are BSNJ

Number of arrangement in this scenerio is 5!= 120

But we also know that each of the vowels can swap positions in this arrangement. Each one can maintain 4 different positions. Thus, the the number of ways of arrangement if the vowels must be together is 120 x 4 =480 ways

Finally, the number of ways of arranging the name OBASANJO if the vowels must not be together = Total number of arrangements possible - Total number of arrangements when vowels are together

This implies 1680-480
Therefore, answer = 1200 ways

I hope this helps?
Nice one Naturalwave
Re: Nairaland Mathematics Clinic by naturalwaves: 10:10pm On Nov 18, 2019
Mrshape:

Nice one Naturalwave
Thanks bro.
Been a while.
I have been damn too busy.
Re: Nairaland Mathematics Clinic by Akanoo1: 4:47am On Nov 19, 2019
naturalwaves:


Step 1.

You have to find the total number of ways the entire words can be arranged and that is 8!/(2! 2!)- This is because letters O and A are doubled.

8!/(2!2!)= 1680 ways


Step 2

You have to find the number of ways the word can be arranged if the vowels must be together. E. G
(OAAO) BSNJ.

This implies that the vowels are 1 letter and we have 4 other letters which are BSNJ

Number of arrangement in this scenerio is 5!= 120

But we also know that each of the vowels can swap positions in this arrangement. Each one can maintain 4 different positions. Thus, the the number of ways of arrangement if the vowels must be together is 120 x 4 =480 ways

Finally, the number of ways of arranging the name OBASANJO if the vowels must not be together = Total number of arrangements possible - Total number of arrangements when vowels are together

This implies 1680-480
Therefore, answer = 1200 ways

I hope this helps?
Thank you
Re: Nairaland Mathematics Clinic by naturalwaves: 5:05am On Nov 19, 2019
Akanoo1:
Thank you
You're welcome!
Re: Nairaland Mathematics Clinic by Aybalance: 9:22am On Nov 20, 2019
pls help:
Two drivers Alison and Kelvin are participating in a drag race.Beginning from a standing start they each proceed with a constant acceleration .kelvin covers the last 1/4 of the distance in 3 secs whereas alison covers the last 1/3 of the distance in 4secs .Who wins and by how much time?
Re: Nairaland Mathematics Clinic by naturalwaves: 7:51pm On Nov 21, 2019
Aybalance:
pls help:
Two drivers Alison and Kelvin are participating in a drag race.Beginning from a standing start they each proceed with a constant acceleration .kelvin covers the last 1/4 of the distance in 3 secs whereas alison covers the last 1/3 of the distance in 4secs .Who wins and by how much time?
Will try this when I get back.
Re: Nairaland Mathematics Clinic by Martinez39(m): 10:06pm On Nov 21, 2019
Aybalance:
pls help:
Two drivers Alison and Kelvin are participating in a drag race.Beginning from a standing start they each proceed with a constant acceleration .kelvin covers the last 1/4 of the distance in 3 secs whereas alison covers the last 1/3 of the distance in 4secs .Who wins and by how much time?
Kelvin wins by approximately 0.59435 sec.
Re: Nairaland Mathematics Clinic by Martinez39(m): 10:38pm On Nov 21, 2019
@Aybalance, the solution requires a very basic knowledge of calculus and motion.

KELVIN'S CASE
dv/dx = a (constant acceleration).
By integration,
v = at ———(1) (since v = 0 when t = 0)
Finding the definite integral of (1) from to - 3 to to (last 3 seconds) with respect to t to get the distance of move in the last 3 seconds ie. a quarter of the total distance ¼So . Doing this and solving for So will lead to
So = 12ato - 18a ——(2)
to is the time at which Kelvin crosses the finish line.
If we had integrated (1) and substituted to into the resulting distance function, we would have gotten
So = ½ato² ———(3)

Solving for a in (3) and substituting (2) leads to
to² - 24to + 36 = 0
Solving this quadratic equation will yield
to ≈ 22.392304 sec
The other solution is less than 3 but we discard it since Kelvin's journey took more than 3 sec.


ALISON'S CASE
Your basically follow the same mathematical procedure in Kelvin's case. The difference is that you integrate from to - 4 to to (last 3 seconds), where to is the time at which Alison crossed the finish line. After everything, you will obtain the quadratic equation
to² - 24to + 48 = 0
The solution is to ≈ 21.797959
We discard the other solution because it is less that 4 seconds and we know that Alison's journey took more than 4 seconds.


Kelvin wins by approximately 0.59435 sec. grin

3 Likes

Re: Nairaland Mathematics Clinic by Aybalance: 6:20pm On Nov 26, 2019
Thanks man....I saw the question under the topic differential equations ....but seems ur approach also works!!!

1 Like

Re: Nairaland Mathematics Clinic by Martinez39(m): 7:10pm On Nov 26, 2019
Aybalance:
Thanks man....I saw the question under the topic differential equations ....but seems ur approach also works!!!
Solving differential equations involves integration. I used integration. Did my answer accord with that given in your textbook?
Re: Nairaland Mathematics Clinic by Brunicekid(m): 7:54pm On Nov 26, 2019
Billyonaire369:
pls solve for me
OK...Solution to this is on its way.
Re: Nairaland Mathematics Clinic by Aybalance: 11:44am On Nov 28, 2019
Martinez39:
Solving differential equations involves integration. I used integration. Did my answer accord with that given in your textbook?
Sure it does!!!approximately 0.594 secs

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