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September 30, 2025, 08:43:15 am

Author Topic: hielly's maths thread  (Read 24948 times)  Share 

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Hielly

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hielly's maths thread
« on: December 23, 2008, 03:53:03 pm »
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hey
im going to post up questions im having difficulties in this thread

provide working out

thanks
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ell

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Re: just a linear problem
« Reply #1 on: December 23, 2008, 04:00:09 pm »
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Using information given:















Substituting x back into the original equations, the first boy has 20 marbles, the second 34, and the third 17.

bec

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Re: Douche's linear problems!
« Reply #2 on: December 25, 2008, 07:56:18 pm »
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Not sure about this (didn't do spesh) but could the equations be:





You need 252L of the 40% solution and 448L of the 15% solution.

Hope that helps/is correct!

Hielly

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Re: hielly's linear problems!
« Reply #3 on: January 12, 2009, 10:25:30 pm »
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okay here are some that i need help.

1) an investor received $31 000 interest per annum from a sum of money, with part of it invested 10% and the remainder at 7% simple interest. she found that if she interchanged the amounts she had invested she could increase her return by $1000 per annum. calculate the total amount she had invested

2) a biathlon event involves running and cycling. kim can cycle 30 km/h faster than she can run. if kim spends 48 minutes running and a third as much time again cycling in an event that covers a total distance of 60 km, how fast can she run?

3)solve the following linear equation.
-2(5-x)/8 + 6/7 = 4(x-2)/3


thanks

Flaming_Arrow

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Re: hielly's linear problems!
« Reply #4 on: January 12, 2009, 11:48:43 pm »
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2.  run = x          cycle = x + 30

where x is the speed

distance = speed * time

 

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Flaming_Arrow

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Re: hielly's linear problems!
« Reply #5 on: January 12, 2009, 11:57:13 pm »
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get a common denominator in one side





cross multiply







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Flaming_Arrow

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Re: hielly's linear problems!
« Reply #6 on: January 13, 2009, 02:20:37 am »
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Let x be the part invested.
Let y be the simple interest.
















therefore $370,588.24 is total amount she had invested
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Hielly

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Re: hielly's linear problems!
« Reply #7 on: January 13, 2009, 11:26:54 am »
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hey thanks flaming arrow, for question 2 the answer in the book is 15km/h

Flaming_Arrow

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Re: hielly's linear problems!
« Reply #8 on: January 13, 2009, 12:00:25 pm »
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hey thanks flaming arrow, for question 2 the answer in the book is 15km/h

but that would mean the total distance is 24km

15 * 48/60 + 45*16/60 = 24km
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Hielly

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Re: hielly's linear problems!
« Reply #9 on: January 13, 2009, 12:39:41 pm »
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hrm yerh the book must have mistaken it. also, the wording for this part is a bit confusing can you please explain this

if kim spends 48 minutes running and a third as much time again cycling in an event that covers a total distance of 60 km

thanks

Flaming_Arrow

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Re: hielly's linear problems!
« Reply #10 on: January 13, 2009, 12:41:32 pm »
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it means if she spends x time running, she spent x/3 times cycling
« Last Edit: January 13, 2009, 12:59:54 pm by Flaming_Arrow »
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Flaming_Arrow

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Re: hielly's linear problems!
« Reply #11 on: January 13, 2009, 04:25:58 pm »
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hey if we substituted it,
say this is equation 1.
x+x+30=60

thus 2x+3-=60 (1)
lets say x=15 what the book said
then,
15+15+30=60
60=60. correct
pretty sure you have to find 2 equations and then use elimination or substitution method.



that doesn't make sense x is the speed and 60 is the distance so how can x + x + 30 = 60?

distance = speed * time
« Last Edit: January 13, 2009, 04:33:35 pm by Flaming_Arrow »
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Hielly

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Re: hielly's linear problems!
« Reply #12 on: January 14, 2009, 12:02:17 am »
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jrm was just trying to see of 15 km/h is actually right but nvm. then the book is wrong .

Flaming_Arrow

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Re: hielly's linear problems!
« Reply #13 on: January 14, 2009, 12:17:38 am »
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Flaming_Arrow

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Re: hielly's linear problems!
« Reply #14 on: January 14, 2009, 12:27:40 am »
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how did you gt to the 6 ?


since they have the same power we times the bases
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