Answer:
The answers to the problem are as follows:
a.) p=\frac{kqr^2}{s}p=
s
kqr
2
b.) k=3k=3
c.) p=96p=96
d.) s=6s=6
Step-by-step explanation:
Variation is a relationship between two distinct objects. There are two types of variation. The first one is called Direct Variation. Direct variation happens when two variables go in the same direction. Furthermore, this happens when one variable increases, the other one increases too. The second type is the Inverse Variation. Inverse variation happens when the two variables go in the opposite direction. Furthermore, this happens when one variable increases, the other one decreases.
Two Types of Variation
1. Direct Variation or Directly proportional
2. Inverse Variation or Indirectly proportional
In mathematical definition, when we say yy varies directly as xx , then there exists a kk such that y=kxy=kx . On the other hand, when we say
Steps in Solving the Problem
1. Write the correct equation. Combined variation problems are solved using a combination of direct variation and inverse variation.
2. Use the information given in the problem to find the value of kk , called the constant of variation or the constant of proportionality.
3. Solve the unknown variable using the equation from step 1 and by substituting the value of kk .
a.) Since pp varies directly as qq and r^2r
2
, and inversely as ss , then our equation must be p=\frac{kqr^2}{s}p=
s
kqr
2
.
b.) Using the equation in part (a), substitute the values of p=40p=40 , q=5q=5 , r=4r=4 , and s=6s=6 into the equation and solve for kk .
\begin{gathered}\begin{aligne}p&=\frac{kqr^2}{s}\\40&=\frac{k(5)(4^2)}{6}\\40&=\frac{80k}{6}\\40&=\frac{40k}{3}\\(3)40&=(3)\left(\frac{40k}{3}\right)\\120&=40k\\\frac{120}{40}&=\frac{40k}{40}\\3&=k\end{aligned}\end{gathered}
c.) Using the value of k=3k=3 and the equation in (a), substitute the value of q=8q=8 , r=6r=6 , and s=9s=9 into the equation and solve for pp .
\begin{gathered}\begin{aligned}p&=\frac{kqr^2}{s}\\&=\frac{3(8)(6^2)}{9}\\&=\frac{864}{9}\\&=96\\\end{aligned}\end{gathered}
p
=
s
kqr
2
=
9
3(8)(6
2
)
=
9
864
=96
d.) Using the value of k=3k=3 and the equation in (a), substitute the value of p=10p=10 , q=5q=5 and r=2r=2 into the equation and solve for ss .
\begin{gathered}\begin{aligned}p&=\frac{kqr^2}{s}\\10&=\frac{(3)(5)(2^2)}{s}\\10&=\frac{60}{s}\\s(10)&=s\left(\frac{60}{s}\right)\\10s&=60\\\frac{10s}{10}&=\frac{60}{10}\\s&=6\end{aligned}\end{gathered}
p
10
10
s(10)
10s
10
10s
s
=
s
kqr
2
=
s
(3)(5)(2
2
)
=
s
60
=s(
s
60
)
=60
=
10
60
=6
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