Mamagogoidn
Answered

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Solve the following:
If y varies directly with the square of x and inversely with z, and y =1 when x=2 and z=10. Find y when x=4 and z=5.​

Sagot :

Hajrudinidn

Answer:

8

Step-by-step explanation:

y = kx²/z

1 = k2²/10

1 = k4/10

10 = k4

k = 10/4

y = 10/4(4)²/5

y = 8

RobbieRavenclawidn

Answer:

The value of y is 8.

Step-by-step explanation:

Mathematical Equation:

[tex]\LARGE\text{$y=\frac{kx^2}{z}$}[/tex]

For the constant of variation (k):

Given:   [tex]y=1[/tex],   [tex]x=2[/tex],   [tex]z=10[/tex]

Find:   [tex]k=?[/tex]

Formula:   [tex]y=\frac{kx^2}{k}[/tex]

Solution:

[tex]y=\frac{kx^2}{y}\\1=\frac{(2)^2k}{10}\\1=\frac{4k}{10}\\4k=(10)(1)\\4k=10\\\frac{4k}{4}=\frac{10}{4}\\k=\frac{10}{4}\\\boldsymbol{k=\frac{5}{2}}[/tex]

For the value of y:

Given:   [tex]k=\frac{5}{2}[/tex],   [tex]x=4[/tex],   [tex]z=5[/tex]

Find:   [tex]y=?[/tex]

Formula:   [tex]y=\frac{kx^2}{z}[/tex]

Solution:

[tex]y=\frac{kx^2}{z}\\y=\frac{(\frac{5}{2})(4)^2}{5}\\y=\frac{(\frac{5}{2})(16)}{5}\\y=\frac{(5)(8)}{5}\\\boxed{\boldsymbol{y=8}}[/tex]

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