Re: [Math4u] got some math I need help with

Problem 1

Without thinking in advance, we just convert the words into math equations until we have enough information to solve the problem.

W = the amount of fuel used each week

S = the amount on hand = 8*W

S=6*(W+20,000 liters)

Now we have enough information to solve for W, probably 3 different ways.

Method 1

Subtract the second equation from the first equation

S-S=8W-6(W+20,000)

0=2W-120,000

120,000 liters = 2W

60,000 liters = W

480,000 liters = 8W = S answer

Method 2

S=8W=6(W+20,000)

++8W=6(W+20,000)

2W=120,000

W=60,000

S=8W=480,000 liters (answer)

Method 3

Use matrices. I didn't pay attention when they taught this.

Problem 2

Method 1

This is bad but it is the intuitive method I would have used before I studied engineering

(50-20)/(100-20)*10=(3/8)*10=3.75

Method 2 Logical method

X = number of liters replaced

(10-X)=number of old liters

0.2*(10-X)=liters of old glycol

1.0*X=liters of new glycol

0.2*(10-X)+1.0*X=total liters of resulting glycol=5

2- 0.2X + 1.0X = 5

0.8X = 3

X=3/0.8=3.75

Method 3

+++++ total +++ glycol +++ % glycol

begin ++10 +++ +++2 +++++ 20

add +++X +++++++X +++++100

result (10+X) ++++(2+X) +++++ 50

(2+X) = 0.5*(10+X)

2+X=5+0.5X

X=6

But now we have 16 liters of solution. So for 10 liters of solution, we need 6*(10/16)=3.75 liters to be replaced.

Problem 3

First you calculate 3pi/5. You could put this into a calculator. Otherwise this is past 90 degrees or past pi/2 so the cosine is negative. Calculate (3pi/5-pi/2). Look up the cosine and put a minus sign in front of it.

Regards,

Brian

 

 

 

 

 

----- Original Message -----

From: Tim Johnson

To: Math4u@yahoogroups.com

Sent: Sunday, November 04, 2007 9:54 AM

Subject: [Math4u] got some math I need help with

I'm a Mature college student that is not getting the math

Here are 3 ? I have answers but need the formulas

#1: A fuel storage depot had an 8-weeks supply on hand. However, cold weather caused the supply to be used in 6 weeks when 20000l extra were used each week. How many liters were in the original supply.    (answer: 480 000)

So X= the total supply       

#2 : A 10 l cooling system contains a solution of 20% antifreeze and 80% water. How much solution must be replaced with pure Antifreeze to raise the strength to 50% antifreeze.   (answer is 3.75L)

#3 cos(3pi/5)    = -.309017

Any help or direction would be great

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