Balansaglloydfruelanidn
Answered

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EVALUATION:
Direction: Answer the following question. Show your solution.
1.Given that y varies directly as x and y = 56 when x=7, find the constant of variation.
2.If y varies directly as x and y=6 when x = 4, find y when x = 10
3. y varies inversely as x and y = 50 when x=4, find y when x = 40.
4.w varies directly as the square of x and inversely as y and z. If w= 12 when x = 4, y = 2 and z=20, find w when x =
3, y = 8 and z= 5.
5.z varies jointly as x and y. If z = 3 when x = 3 and y = 15, find z when x = 6 and y = 9.
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Sagot :

Jamelbenibrahimidn

Answer:

[tex]1. \: y = kx \\ 56 = k \times 7 \\ \frac{56}{7} = k \\ k = 8 \: (constant \: of \: variation)[/tex]

[tex]2. \: y = kx \\ 6 = k \times 4 \\ \frac{6}{4} = k \\ k = \frac{3}{2} (constant \: of \: variation)[/tex]

hi[tex]y = kx \: \: - > \: \: y = \frac{3}{2} x \\ when \: x = 10 \\ y = \frac{3}{2} \times 10 = 15 \\ y = 15[/tex]

[tex]3. \: y = \frac{k}{x} \\ 50 = \frac{k}{4} \\ 50 \times 4 = k \\ k = 200 \: (constant \: of \: variation)[/tex]

[tex]y = \frac{k}{x} = \frac{200}{x} \\ when \: x = 40 \\ y = \frac{200}{40} = 5 \\ y = 5[/tex]

[tex]4. \: w = \frac{k {x}^{2} }{yz} \\ 12 = \frac{k \times {4}^{2} }{2 \times 20} \\ 12 = \frac{16k}{40} \\ \frac{12 \times 40}{16} = k \\ k = 30 \: (constant \: of \: variation)[/tex]

[tex]w = \frac{k {x}^{2} }{y} \: \: - > \: \: w = \frac{200 {x}^{2} }{yz} \\ when \: x = 3 \: y = 8 \: z = 5 \\ w = \frac{200 \times {3}^{2} }{8 \times 5} = \frac{200 \times 9}{40} = 45 \\ w = 45[/tex]

[tex]5. \: z =k xy \\ 3 = k \times 3 \times 15 \\ \frac{3}{3 \times 15} = k \\ k = \frac{1}{15} (constant \: of \: variation)[/tex]

[tex]z = kxy \: \: - > \: \: z = \frac{1}{15} xy \\ when \: x = 6 \: \: y = 9 \\ z = \frac{1}{15} \times 6 \times 9 = \frac{54}{15} = \frac{18}{5} = 3 \frac{3}{5} \\ z = 3 \frac{3}{5}[/tex]

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