Respuesta :
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[tex] \large \sf \underline{Problem:}[/tex]
Its volume V. If r varies directly as s and inversly as t, r=27, when s=18 and t=2 find:
- 1.) r when t=3 and s=27
- 2.) s when t=2 and r=3
- 3.) t when r=1 and s=6
- 4.) r when s=4 and 1=2
- 5.) s when t=5 and r=6
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[tex] \large \sf \underline{Answers:}[/tex]
[tex] \qquad \quad \huge \sf{1. r= 27} \\ \\ \qquad \huge \sf{2. s= 2 } \\ \\ \qquad \quad \huge \sf{3. t = 18} \\ \\ \qquad \huge \sf{4. r = 6 } \\ \\ \qquad \quad \huge \sf{5. s = 10}[/tex]
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[tex] \large \sf \underline{Solution:}[/tex]
Combined Variation:
[tex] \large\bold{r=\frac{ks}{t}}\:\:,\:\:\sf s=\frac{rt}{k}\:\:,\:\:\sf t=\frac{ks}{r}\:\:,\:\:\sf k=\frac{rt}{s}[/tex]
Given:
- r = 27
- s = 18
- t = 2
Find the constant (k)
[tex] \begin{gathered}\begin{aligned}&\sf k=\frac{rt}{s}\\&\sf k=\frac{27(2)}{18}\\&\sf k=\frac{54}{18}\\&\sf k=3\end{aligned}\end{gathered} [/tex]
Use k = 3 to solve the following
Number 1:
[tex]\begin{gathered}\begin{aligned}&\sf r=\frac{ks}{t}\\&\sf r=\frac{\cancel3\times 27}{\cancel3}\\&\underline{\bold{\pmb{r=27}}}\end{aligned}\end{gathered} [/tex]
Number 2:
[tex]\begin{gathered}\begin{aligned}&\sf s=\frac{rt}{k}\\&\sf s=\frac{\cancel3\times 2}{\cancel3}\\&\underline{\bold{\pmb{s=2}}}\end{aligned}\end{gathered} [/tex]
Number 3:
[tex]\begin{gathered}\begin{aligned}&\sf t=\frac{ks}{r}\\&\sf t=\frac{3\times 6}{1}\\&\underline{\bold{\pmb{t=18}}}\end{aligned}\end{gathered} [/tex]
Number 4:
[tex]\begin{gathered}\begin{aligned}&\sf r=\frac{ks}{t}\\&\sf r=\frac{3\times \cancel4}{\cancel2}\\&\sf r=3\times 2\\&\underline{\bold{\pmb{r=6}}}\end{aligned}\end{gathered} [/tex]
Number 5:
[tex]\begin{gathered}\begin{aligned}&\sf s=\frac{rt}{k}\\&\sf s=\frac{\cancel6 \times 5}{\cancel3}\\&\sf s=2\times 5\\&\underline{\bold{\pmb{s=10}}}\end{aligned}\end{gathered} [/tex]
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#LetEarthBreathe
[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]
According to the question :
[tex]\qquad \sf \dashrightarrow \:r \propto s[/tex]
and
[tex]\qquad \sf \dashrightarrow \:r \propto \dfrac{1}{t} [/tex]
Now, by both equations we can infer that ~
[tex]\qquad \sf \dashrightarrow \:r \propto \dfrac{s}{t} [/tex]
now, assume k to be a proportionality constant
[tex]\qquad \sf \dashrightarrow \:r = k \cdot \dfrac{s}{t} [/tex]
Now, plug in the value of given values, to find value of k ~
[tex]\qquad \sf \dashrightarrow \:27 = \dfrac{18}{2} \cdot k[/tex]
[tex]\qquad \sf \dashrightarrow \:27 = 9{} k[/tex]
[tex]\qquad \sf \dashrightarrow \:k = 27 \div 9[/tex]
[tex]\qquad \sf \dashrightarrow \:k = 3[/tex]
Now, let's evaluate the required values ~
# Question 1
[tex]\qquad \sf \dashrightarrow \:r = 3 \cdot \dfrac{s}{t} [/tex]
[tex]\qquad \sf \dashrightarrow \:r = 3 \cdot \dfrac{27}{ 3} [/tex]
[tex]\qquad \sf \dashrightarrow \:r = 27[/tex]
# Question 2
[tex]\qquad \sf \dashrightarrow \:r= 3 \cdot \dfrac{s}{t} [/tex]
[tex]\qquad \sf \dashrightarrow \:3= 3 \cdot \dfrac{s}{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:{s}{} = \dfrac{3 \times 2}{3} [/tex]
[tex]\qquad \sf \dashrightarrow \:s = 2[/tex]
# Question 3
[tex]\qquad \sf \dashrightarrow \:r= 3 \cdot \dfrac{s}{t} [/tex]
[tex]\qquad \sf \dashrightarrow \:1= 3 \cdot \dfrac{6}{t} [/tex]
[tex]\qquad \sf \dashrightarrow \:t = 6\cdot3[/tex]
[tex]\qquad \sf \dashrightarrow \:t = 18[/tex]
# Question 4
[tex]\qquad \sf \dashrightarrow \:r = 3 \cdot \dfrac{s}{t} [/tex]
[tex]\qquad \sf \dashrightarrow \:r = 3 \cdot \dfrac{4}{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:r = \dfrac{12}{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:r =6[/tex]
# Question 5
[tex]\qquad \sf \dashrightarrow \:r = 3 \cdot \dfrac{s}{t} [/tex]
[tex]\qquad \sf \dashrightarrow \:6= 3 \cdot \dfrac{s}{5} [/tex]
[tex]\qquad \sf \dashrightarrow \:s = \dfrac{6 \times 5}{3} [/tex]
[tex]\qquad \sf \dashrightarrow \:s = 10[/tex]
