Question 59355
Find an equation relating the variables for each situation described.
:
(1) z varies directly as b, and z = 4 when b = 16
  "z varies direclty as b" -->> z=kb
z=4 and b=16 find k:
4=k(16)
{{{4/16=16k/16}}}
{{{1/4=k}}}
Substitute that into the first step:
{{{highlight(z=(1/4)b)}}}
:

(2) s2 varies inversely as s1, and s2= -2 when s1 = 1/2
"s2 varies inversely as s1" -->> s2=k/s1
s2=-2 and s1=1/2 find k:
{{{-2=k/(1/2)}}}
{{{(1/2)(-2)=(1/2)k/(1/2)}}}
{{{-1=k}}}
Substitute that into the first step:
{{{highlight(s2=-1/s1)}}}
:
(3) y varies jointly as a and b, and y = -24 when a = 9 and b = -4
"y varies jointly as a and b" -->> y=kab
y=-24, a=9, and b=-4, find k
-24=k(9)(-4)
-24=-36k
-24/-36=-36k/-36
2/3=k
Substitute that into the first step:
{{{highlight(y=(2/3)ab)}}}
:
(4) m varies directly as n and inversely as p, and p = -3 and m = 1 and n = 1/2
"m varies directly as n and inversely as p" -->> m=kn/p
p=-3, m=1, and n=1/2 find k
{{{1=k(1/2)/-3}}}
{{{-3(1)=-3k(1/2)/-3}}}
{{{-3=k/2}}}
{{{2(-3)=2k/2}}}
{{{-6=k}}}
Substitute that into step 1:
{{{highlight(m=-6n/p)}}}
:
(5) The cost of pipe is directly proportional to its length.  If 6 feet of pipe costs $40, find the cost of 10 feet of pipe.
Let cost be:C  and Length be:L
"cost is directly proportional to length" -->> C=kL
C=40, L=6, find k:
40=k(6)
40/6=6k/6
6.667=k
Substitute that into the first step: {{{C=(6.667)L}}}
Let L=10
C=(6.667)(10)
C=$66.67
:
(6) The intensity of light is inversely prorportional to the square of its distance from its origin. An intensity of 100 lumens is registered at a distance of 5 feet.  What is the intensity of light registered at a distance of 10 feet?
Let intensity be:I and distance be:d
"intensity of light in inversely proportional to the square of its distance" -->> {{{I=k/d^2}}}
I=100, d=5, find k
{{{100=k/(5^2)}}}
{{{100=k/25}}}
{{{25(100)=25k/25}}}
{{{2500=k}}}
Substitute that into step 1:
{{{I=2500/d^2}}}
If d=10
{{{I=2500/(10^2)}}}
{{{I=2500/100}}}
I=25 lumens
Happy Calculating!!!