Re: variation

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Posted by Joel on October 04, 2002 at 12:18:48:

In Reply to: variation posted by rk on October 04, 2002 at 10:11:25:

: Can u please provide step by step solutions

: 1/ If y varies partly directly as x and partly inversely as x, and y=2 1/2 when x=1, and y=2 when x=2, find an equation
: connecting the two varibles.

: 2/ y is the sum of two quantities, one of which varies as x and the other x squared. When x=4, y=16, and when x=6,
: y=30.Express y in terms of x.

: 3/ Two varibles x and y are related by the formula y=ax+b/x squared. If y=1 1/5 when x=5 and y=2 1/20 when x=10, find
: the value of y when x =15

: 4/ The cost, C, of publishing a bulletin consists of two parts:
: -a fixed charge,a,for setting up the type, independent of the number of copies published
: - a charge for printing and paper,proportional to the number of copies printed,n. If the cost of 500 copies is \$20 and the
: cost of 750 copies is \$25,find the cost of 900 copies.

I'll get you started:
1) This problem doesn't explicitly state that y is the sum of the two quantities but I will assume that that is what was intended. Take two constants a and b. Then:
y = a*x + b*(1/x)
y = a*x + b/x Equation 1
Next substitute the given values of x and y to get two equations with two unknowns:
5/2 = a*1 + b/1
2 = a*2 + b/2
Solve for a and b, then plug those values into Equation 1.

2) This is almost the same as #1; here the general equation is
y = a*x + b*x^2 Equation 2
Again, substitute the given values of x and y:
16 = a*4 + b*16 {note 4^2 = 16}
30 = a*6 + b*36 {note 6^2 = 36}
Now, solve for a and b and plug them into Equation 2.

3)You have to write this more clearly. What is squared (ax + b/x)^2 or ax + (b/x)^2 or ax + b/(x^2). Anyway, once you determine that, the solution works the same as #1 and #2, so why don't you see what you can do yourself?

4)Same process. Give it a try & let's see what you can do with it.

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