saipn:
Elders in the house help me look into this questions

QUESTION 1
In the first sample, to get the top right number,
You add whatever number you have on the bottom right to the number that is opposite it(top left) and then you multiply by the bottom right number again.
From 1st diagram,
bottom right = 3
number opposite(top left) = 1
3+1=4
now, this 4 will be multiplied by bottom right(3) to get top right = 12

Again, (7+2) x7 = 63
and again, (5+5) x 5 = 50

To get the bottom left number,
You do (bottom right) - (top left)
Whatever number you get, you then multiply by itself to get bottom left
E.g 3-1= 2
2 x 2= 4
In second one,
7- 2= 5
5x 5= 25
in third one
5-5 = 0
0 x0 = 0
Solution to given question
(X+3)x X= 40
X(X+3)=40
X²+3X=40
X= 5 BECAUSE 25 +15= 40

Answer= 5

QUESTION 2
This one is straight forward. 2 numbers at the bottom when multiplied will give top left
X x 3 = 21
x= 7
Answer= 7

naturalwaves:

QUESTION 1
In the first sample, to get the top right number,
You add whatever number you have on the bottom right to the number that is opposite it(top left) and then you multiply by the bottom right number again.
From 1st diagram,
bottom right = 3
number opposite(top left) = 1
3+1=4
now, this 4 will be multiplied by bottom right(3) to get top right = 12

Again, (7+2) x7 = 63
and again, (5+5) x 5 = 50

To get the bottom left number,
You do (bottom right) - (top left)
Whatever number you get, you then multiply by itself to get bottom left
E.g 3-1= 2
2 x 2= 4
In second one,
7- 2= 5
5x 5= 25
in third one
5-5 = 0
0 x0 = 0
Solution to given question
(X+3)x X= 40
X(X+3)=40
X²+3X=40
X= 5 BECAUSE 25 +15= 40

Answer= 5

QUESTION 2
This one is straight forward. 2 numbers at the bottom when multiplied will give top left
X x 3 = 21
x= 7
Answer= 7

Chai God will surely bless you thanks boss

saipn:
Chai God will surely bless you thanks boss

Amen o.

``If 4/3, m,1,n....... Is GP, the product of m and n is ? I just need the workings ooh

Akahmech:

``If 4/3, m,1,n....... Is GP, the product of m and n is ? I just need the workings ooh

for any G.P, the common ratio is given as T(n+1)/T(n) I.e
r = T2/T1 = T3/T2 = T4/53 = . . .
In this particular case, r = m/(4/3) = 1/m = n/1
The equality that concerns us is the last one
1/m = n/1
From which cross multiplication gives the product mn = 1

Where is Richard? Engineer where you dey?

Pls some one should help me find this limit.

mathefaro:
for any G.P, the common ratio is given as T(n+1)/T(n) I.e
r = T2/T1 = T3/T2 = T4/53 = . . .
In this particular case, r = m/(4/3) = 1/m = n/1
The equality that concerns us is the last one
1/m = n/1
From which cross multiplication gives the product mn = 1

thanks sir

naturalwaves:


Amen o.

Happy Sunday boss pls help me to view this question and the method to use

saipn:
Happy Sunday boss pls help me to view this question and the method to use

the answer is 37 ,(( 5*10)+8 ) -21

saipn:
Happy Sunday boss pls help me to view this question and the method to use

Just saw this bro. Someone has solved it and his solution is correct. I just want you to improve your hand writing. Take your time to write your figures properly.

naturalwaves:

Just saw this bro. Someone has solved it and his solution is correct. I just want you to improve your hand writing. Take your time to write your figures properly.

ok thanks

Electricboy:
the answer is 37 ,(( 5*10)+8 ) -21

Thank you

A mans salary in 1975 was N2000.00 per year and it increased by 10% each year. Find how much he earned in the years up to 1984 both years inclusive, please help , word problems on sequence and series

The salary in 1975, 1976, 1977... will form a GP with a = 2000, r = 1.1 (2000, 2200, 2420, 2662...)

Sum of the terms from 1775 to 1984 (ten years, since both years are inclusive)
So n = 10. S = a(r^n-1)/r-1
S = 2000 (1.1^10 - 1)/(1.1-1)
= N31874.85

Akahmech:
A mans salary in 1975 was N2000.00 per year and it increased by 10% each year. Find how much he earned in the years up to 1984 both years inclusive, please help , word problems on sequence and series

Thanks ooh I will send more questions

doggedemmy:
The salary in 1975, 1976, 1977... will form a GP with a = 2000, r = 1.1 (2000, 2200, 2420, 2662...)

Sum of the terms from 1775 to 1984 (ten years, since both years are inclusive)
So n = 10. S = a(r^n-1)/r-1
S = 2000 (1.1^9 - 1)/(1.1-1)
= N27158.95

Akahmech:
A mans salary in 1975 was N2000.00 per year and it increased by 10% each year. Find how much he earned in the years up to 1984 both years inclusive, please help , word problems on sequence and series

solve the problem, I used Excel in my computer.


The printout is this Table below


1975 2000.00
1976 2200.00
1977 2420.00
1978 2662.00
1979 2928.20
1980 3221.02
1981 3543.12
1982 3897.43
1983 4287.18
1984 4715.90


the sum 31874.85

Akahmech:
Thanks ooh I will send more questions

Check, I just modified my post.

Nice job, please where is mathschic?

Nasinazz:
Nice job, please where is mathschic?

You just mentioned my crush anyhow

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

PLEASE Help me out
An aeroplane flies from a town X on a bearing of N45E to another town Y, a distance of 200km. It then changes course and flies to another town Z on a bearing of S60E. If Z is directly east of X, calculate correct to 3 significant figures,
a. the distance from X to Z
b. the distance from Y to XZ

Thanks.

pemisire017:


PLEASE Help me out
An aeroplane flies from a town X on a bearing of N45E to another town Y, a distance of 200km. It then changes course and flies to another town Z on a bearing of S60E. If Z is directly east of X, calculate correct to 3 significant figures.
Thanks.

You haven't stated what to calculate

pemisire017:


PLEASE Help me out
An aeroplane flies from a town X on a bearing of N45E to another town Y, a distance of 200km. It then changes course and flies to another town Z on a bearing of S60E. If Z is directly east of X, calculate correct to 3 significant figures.
Thanks.

incomplete

mathefaro:
You haven't stated what to calculate

a. the distance from X to Z
b. the distance from Y to XZ

Thanks

pemisire017:


a. the distance from X to Z
b. the distance from Y to XZ

Thanks

here

pemisire017:


a. the distance from X to Z
b. the distance from Y to XZ

Thanks

bros thankGod i found you.

Ive been trying to teach my younger bro further maths recently and ive been having difficulties.

I dont know if im also free to post questions here too?

pemisire017:


a. the distance from X to Z
b. the distance from Y to XZ

Thanks

Draw the diagram and you'll get a triangle XYZ with the angles

YXZ = 45°

ZYX = 105° (60 + 45)

XZY = 30° ( 180-105 - 45)

Then you can find XZ by the sine rule:

XZ/sin 105 = 200/sin 30

XZ = 200 sin105/ sin 30

XZ = 386.37 km.

Similarly for YZ

YZ/sin45 = 200/sin 30

YZ = 200 sin45/sin 30

YZ = 282.84 km.

Alhajiemeritus:


Draw the diagram and you'll get a triangle XYZ with the angles

YXZ = 45°

ZYX = 105° (60 + 45)

XZY = 30° ( 180-105 - 45)

Then you can find XZ by the sine rule:

XZ/sin 105 = 200/sin 30

XZ = 200 sin105/ sin 30

XZ = 386.37 km.

Similarly for YZ

YZ/sin45 = 200/sin 30

YZ = 200 sin45/sin 30

YZ = 282.84 km.

The question didn't ask for YZ, check again

jumper524:
bros thankGod i found you.

Ive been trying to teach my younger bro further maths recently and ive been having difficulties.

I dont know if im also free to post questions here too?

Why not? We have people here who are willing to render assistance

jumper524:
bros thankGod i found you.

Ive been trying to teach my younger bro further maths recently and ive been having difficulties.

I dont know if im also free to post questions here too?

Hahaha
Welcome to the house.

mathefaro:
here

Your handwriting is beautiful.