COMBINED VARIATION
derive the statement of
variation into mathematical
equation
solve problems involving
combined variation
apply the concept of
OBJECTIVES
combined variation in various
real - life situations
COMBINED VARIATION
Combined Variation
the statement y varies
when one quantity varies directly as “x and inversely
directly with one or more as y” in symbol:
variables and inversely
with one ore more y=
variables where k is the constant of
variation
Translate the following
statement into mathematical
equation
Translate the following statement
into mathematical equation
1. t varies directly as a and inversely as b.
2. y varies directly as x and inversely as the square of z
3. P varies directly as the square of x and inversely as s.
4. e varies jointly as f and g and inversely as d.
5. m varies jointly as o and p and inversely as n.
Translate the following statement
into mathematical equation
6. x varies directly as t and w, and inversely as s.
7. the time t required to travel is directly proportional to
temperature t and inversely proportional to the pressure P.
8. the pressure of gas varies directly as its temperature t and
inversely as its volume V.
9. The sales S of a product is directly proportional to the money
M spent on advertising but inversely proportional to its price P
10. Speed S varies directly as the distance traveled d, and varies
inversely as the time t taken.
If z varies directly as x and inversely as y and z = 9 when x
= 6 and y = 2, find z when x = 8 and y = 12
z=
9=
k= z=
z=
k=3
z=2
If r varies directly as s and inversely as the square of u,
and r = 2 when s = 18 and u = 2,
If r varies directly as s and inversely as the square of u,
and r = 2 when s = 18 and u = 2,
r=
2=
2= the equation is r =
=
k=
If r varies directly as s and inversely as the square of u,
and r = 2 when s = 18 and u = 2, find,
r=
find,
a. r when u = 3 and s = 27
b. s when u = 2 and r = 4
c. u when r = 1 and s = 36
If r varies directly as s and inversely as the square of u,
and r = 2 when s = 18 and u = 2, find,
a. r when u = 3 and s = 27
r=
If r varies directly as s and inversely as the square of u,
and r = 2 when s = 18 and u = 2, find,
a. r when u = 3 and s = 27
r=
If r varies directly as s and inversely as the square of u,
and r = 2 when s = 18 and u = 2, find,
a. r when u = 3 and s = 27
r= r=
r=
r=
If r varies directly as s and inversely as the square of u,
and r = 2 when s = 18 and u = 2, find,
b. s when u = 2 and r = 4
If r varies directly as s and inversely as the square of u,
and r = 2 when s = 18 and u = 2, find,
b. s when u = 2 and r = 4
r= 16 =
4= 144 = 4s
4= =
s = 36
If r varies directly as s and inversely as the square of u,
and r = 2 when s = 18 and u = 2, find,
c. u when r = 1 and s = 36
If r varies directly as s and inversely as the square of u,
and r = 2 when s = 18 and u = 2, find,
c. u when r = 1 and s = 36
r= 1=
1=
= 16
u=4
The maximum load m of a beam varies directly as the breadth b and the
square of the depth d and inversely as the length l. If a beam is 3.6
meters long, 1.2 meters wide, and 2.4 meters deep can safely bear a
load uo to 909kg, find the maximum safe load for a beam of the same
material which is 3m long, 6m wide and 1.8m deep.
The maximum load m of a beam varies directly as the breadth b and the
square of the depth d and inversely as the length l. If a beam is 3.6
meters long, 1.2 meters wide, and 2.4 meters deep can safely bear a
load uo to 909kg, find the maximum safe load for a beam of the same
material which is 3m long, 6m wide and 1.8m deep.
m= 909 =
therefore the equation
909 = 3272.4 = 6.912k
of variation is
909 = =
473.44 = k m=
The maximum load m of a beam varies directly as the breadth b and the
square of the depth d and inversely as the length l. If a beam is 3.6
meters long, 1.2 meters wide, and 2.4 meters deep can safely bear a
load uo to 909kg, find the maximum safe load for a beam of the same
material which is 3m long, 6m wide and 1.8m deep.
m=
m=
m = 3,067.90kg
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