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Re: Nairaland Mathematics Clinic by Nobody: 3:19am On Apr 01, 2017
1.) Simplify (sqr5+1/sqr5)(sqr5–1/sqr3)
2.)If y= 3x²(x³+1)^½ find dy/dx
3.) solve the inequality 1/3x+5/8>1/2x-2/24 or x/3+5/8>or=x/2-2/24
Re: Nairaland Mathematics Clinic by agentofchange1(m): 3:45pm On Apr 01, 2017
Mrphantom99:
1.) Simplify (sqr5+1/sqr5)(sqr5–1/sqr3)

2.)If y= 3x²(x³+1)^½ find dy/dx

3.) solve the inequality
1/3x+5/8>1/2x-2/24 or x/3+5/8>or=x/2-2/24


Sanp And. Post ...Fast...
Re: Nairaland Mathematics Clinic by Deicide: 5:53pm On Apr 01, 2017
Mrphantom99:
1.) Simplify (sqr5+1/sqr5)(sqr5–1/sqr3)
if its sqr5+1 the answer is
(4sqr15)/(15)
but if it is (sqr(5+1))
the answer is
(2sqr10)/(5)
Re: Nairaland Mathematics Clinic by Darvel(m): 6:46pm On Apr 01, 2017
Pls help me answer dis question... Pls i will appreciate if u solve it somewhere n snap it

Re: Nairaland Mathematics Clinic by agentofchange1(m): 7:57pm On Apr 01, 2017
Here

Re: Nairaland Mathematics Clinic by deeprinz(m): 3:54pm On Apr 02, 2017
Define a norm on any magic square matrix ,and hence show that every magic square Is an inner product space
Re: Nairaland Mathematics Clinic by Nobody: 2:30pm On Apr 03, 2017
An-yeong-ha-se-yo

1 Like

Re: Nairaland Mathematics Clinic by Idenyijoshua(m): 10:29pm On Apr 03, 2017
good evening. please how do I approach this: xy=k, where k is a constant.
Re: Nairaland Mathematics Clinic by profmathsland(m): 12:53am On Apr 09, 2017
Good day to you all great critical thinkers and creative minders. How have we all been? It's been a while though I was on a mathematical journey to an eigenspace where i had to be accommodated by the metric that will be on me soonest. But I am back now. I want us to look deeply at the nature of numbers and its role in Mathematics as a whole because it plays an important role in determining the good and the bad of mathematics. One part of it is in the use of numbers with signs in the popular acronym called BODMAS. It's a pity that when we are taught these things we fail to go back and study them in order to be sure if we are right or wrong. Now two years ago I made a discovery on this acronym called BODMAS and discovered that you can't actually make use of a symbol more than once. The O is the acronym as our teachers will tell us was taken to be Of which is meant to be the sign of multiplication which is also for M. Now I saw that O actually stands for Order which implies powers and roots. The only problem is to gradually put this into correction because we are talking about giving our incoming generations the right message to this subject I call the Queen of all subjects and King of sciences. So what do we think?

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Re: Nairaland Mathematics Clinic by timmykaydude: 12:46pm On Apr 09, 2017
Profmathsland,I have heard that also....nnamdi1998,has your maths questions been answered?if yes,send me the answer on facebook.
Re: Nairaland Mathematics Clinic by profmathsland(m): 7:15pm On Apr 09, 2017
timmykaydude:
Profmathsland,I have heard that also....nnamdi1998,has your maths questions been answered?if yes,send me the answer on facebook.

Which of them please?
Re: Nairaland Mathematics Clinic by Nobody: 5:18pm On Apr 10, 2017
timmykaydude:
nnamdi1998,has your maths questions been answered?.
nope
Re: Nairaland Mathematics Clinic by timmykaydude: 7:22pm On Apr 10, 2017
profmathsland:

Which of them please?
do you mean the mathematics question?
Re: Nairaland Mathematics Clinic by Kentnickole(m): 9:45pm On Apr 10, 2017
Idenyijoshua:
good evening.
please how do I approach this:
xy=k, where k is a constant.

Solution:
given xy=k where k is a constant
It then means that x and y will assume the following values: x=k/y; y=k/x respectively

1 Like

Re: Nairaland Mathematics Clinic by profmathsland(m): 12:25am On Apr 12, 2017
timmykaydude:
do you mean the mathematics question?


Yes
Re: Nairaland Mathematics Clinic by francisdfreak: 12:50pm On Apr 15, 2017
How does one solve an equation which has x as a factor or variable on both sides of the equal sign?
An equation like: 2x=4x, find x or 50x^3/2=32x^-1/2. Find x

A well detailed/step by step solution will be greatly appreciated
Re: Nairaland Mathematics Clinic by Nobody: 1:19pm On Apr 15, 2017
francisdfreak:
How does one solve an equation which has x as a factor or variable on both sides of the equal sign?
An equation like: 2x=4x, find x or 50x^3/2=32x^-1/2. Find x

A well detailed/step by step solution will be greatly appreciated

The first is 2x-4x=0 this implies that -2x=0, hence x=0; the second equation can be solved by multiplying both sides of the equation by x^1/2. So you have 50x^2=32; hence x^2= 32/50; x^2=16/25, which implies x=4/5..
Re: Nairaland Mathematics Clinic by francisdfreak: 1:35pm On Apr 15, 2017
masperano:


The first is 2x-4x=0 this implies that -2x=0, hence x=0; the second equation can be solved by multiplying both sides of the equation by x^1/2. So you have 50x^2=32; hence x^2= 32/50; x^2=16/25, which implies x=4/5..

Thanks a lot bro, i really appreciate.
Looks like i'll be mentioning your moniker anytime i encounter a math problem. grin
Re: Nairaland Mathematics Clinic by Nobody: 2:02pm On Apr 15, 2017
francisdfreak:


Thanks a lot bro, i really appreciate.
Looks like i'll be mentioning your moniker anytime i encounter a math problem. grin

Lol no worries... Cheers!
Re: Nairaland Mathematics Clinic by timmykaydude: 12:41pm On Apr 16, 2017
Re: Nairaland Mathematics Clinic by timmykaydude: 12:53pm On Apr 16, 2017
profmathsland:



Yes
I try to attach the picture,its not showing don't know if its nairaland or my phone...nnamdi1998,attach the question again.
Re: Nairaland Mathematics Clinic by Nobody: 1:49pm On Apr 20, 2017
can you solve this geometry problem for 6th grade in China. grin

Re: Nairaland Mathematics Clinic by naturalwaves: 6:05pm On Apr 20, 2017
masperano:
can you solve this geometry problem for 6th grade in China. grin
I think I will like to give this a try once I am settled.
Re: Nairaland Mathematics Clinic by naturalwaves: 7:23pm On Apr 20, 2017
masperano:
can you solve this geometry problem for 6th grade in China. grin
Area of tringle ACD = 1/2b * h
=1/2 * 20 * 10
= 100
If the 2 different parts of the circle in the down region containing the red areas are joined together, we will get a complete circle. Therefore, we need to subtract the area of the circle from the area of the triangle to get the part that remains.
The radius of the circle is 5 (i.e half way through the side of the rectangle. Hence, area = pie * radius square
= 22/7 * 25
78.57
The remaining portion of the area = 100- 78.57
=21.43
But 21.43 is not the area of the remaining red region. In order to get the area of the red region, we still need to subtract that small part down there around the place I tagged “ A ” and this is where the problem lies.
The final solution will be gotten by subtracting the area of the small region close to where I tagged "A" from 21. 43. I do not know how to get the area of that small region but if I extend the image to make a square and try and see or guess what size of the square that portion will take, I will say around 1/8 of the portion.
Area of square is 25 and 1/8 of it will give 3.125 ( this is just an estimate by eyes o. lol)
THerefore, estimated area of the red region is 21.43- 3.125 = 18.305 metre or centimeter square since you did not give the unit of the dimensions.
Note: The real answer should flabitate between plus or minus 2 from this estimated answer grin

Edited

Calculating the area of the small piece on the left corner
Area of small piece= area of the smaller triangle formed - the right piece - the semi circular part
= 1/2 *10*5 -right piece - semi circular part
= 25 - right piece- semi circular part

Finding the area of the right piece
To get this, we need to extend a square shape to the diagram and subtract a quatre of a circle from it
Area of extended square is 5 square = 25
Area of quatre of circle is (22/7 * 25)/ 4 = 19.64
Area occupied by the right piece = 25- 19.64
= 5.36
So now, we are left with finding just the area of the semi circular part.
NOTE: The semi circle formed by this area was not formed from the bigger triangle, so we cannot use the big triangle to resolve this.

How to solve the area of the semi circular part
The only way to do this is to draw up a triangle on the semi circle such that part of the triangle will serve as the radius of the bigger circle. The reason for doing this is to obtain a sector which we can easily subtract a triangle from in order to get the area of that small part. The triangle formed will be an isosceles triangle.
This will give 17.68 after solving. 17.68 is the area of the semi circular part
Recall that area of the small piece formed = Area of new triangle drawn - area of right piece - area of semi circular part
= 25-5.36-17.68
= 1.96

Therefore, the area of our red region = our initial 21.43- 1.96
= 19.47 metre or centimetre square

Note: I attached the image of how the cut was made and how the area of the semi circular part was resolved. Check image below and please, do not tell me I am wrong embarassed because this question kept me on my sit for 2 hours+ after work shocked shocked. It is so tough o.
Cc: masperano, thankyoujesus, emmitex77,inschool

1 Like

Re: Nairaland Mathematics Clinic by Nobody: 1:44am On Apr 21, 2017
naturalwaves:

Area of tringle ACD = 1/2b * h
=1/2 * 20 * 10
= 100
If the 2 different parts of the circle in the down region containing the red areas are joined together, we will get a complete circle. Therefore, we need to subtract the area of the circle from the area of the triangle to get the part that remains.
The radius of the circle is 5 (i.e half way through the side of the rectangle. Hence, area = pie * radius square
= 22/7 * 25
78.57
The remaining portion of the area = 100- 78.57
=21.43
But 21.43 is not the area of the remaining red region. In order to get the area of the red region, we still need to subtract that small part down there around the place I tagged “ A ” and this is where the problem lies.
The final solution will be gotten by subtracting the area of the small region close to where I tagged "A" from 21. 43. I do not know how to get the area of that small region but if I extend the image to make a square and try and see or guess what size of the square that portion will take, I will say around 1/8 of the portion.
Area of square is 25 and 1/8 of it will give 3.125 ( this is just an estimate by eyes o. lol)
THerefore, estimated area of the red region is 21.43- 3.125 = 18.305 metre or centimeter square since you did not give the unit of the dimensions.
Note: The real answer should flabitate between plus or minus 2 from this estimated answer grin



Lol nice guess.... I still believe if you look at it well you can solve it accurately without eyeballing it. Think about it more and see if you can..I will provide the solution when you demand. cheers mate
Re: Nairaland Mathematics Clinic by naturalwaves: 8:13am On Apr 21, 2017
masperano:


Lol nice guess.... I still believe if you look at it well you can solve it accurately without eyeballing it. Think about it more and see if you can..I will provide the solution when you demand. cheers mate
Okay bro. I think I will need some angle formulas to resolve that o if I wanted to and I have lost touch with most of them. I will try again sha maybe after work but I doubt if this is for 6th grade o.
Re: Nairaland Mathematics Clinic by thankyouJesus(m): 10:47am On Apr 22, 2017
masperano:
can you solve this geometry problem for 6th grade in China. grin
Area of rectangle - 2(area of circle)

you can divide the answer by 8 to get area of each corner......it should be easy from here
Re: Nairaland Mathematics Clinic by Nobody: 11:10am On Apr 22, 2017
thankyouJesus:

Area of rectangle - 2(area of circle)

you can divide the answer by 8 to get area of each corner......it should be easy from here

You forgot that after subtracting the area of the two circles from the area of the rectangle, you are left with areas of the corners of the rectangle and areas above and beneath where the circles are tangent to each other. So not so easy!
Re: Nairaland Mathematics Clinic by Nobody: 11:14am On Apr 22, 2017
naturalwaves:

Okay bro. I think I will need some angle formulas to resolve that o if I wanted to and I have lost touch with most of them. I will try again sha maybe after work but I doubt if this is for 6th grade o.

Lol it is man.. those folks are being trained rigorously when it comes to mathematics from an early stage... believe me.
Re: Nairaland Mathematics Clinic by Emmitex77: 7:56am On Apr 24, 2017
That question is very easy na. . Just use polynomial formula A.K.A factor theorem to solve it.

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