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Marrrta [24]
1 year ago
14

Suppose z varies directly with x and inversely with the square of y. If z = 18 when I = 6 and y = 2, what is z when I 7 and y =

7? Z =

Mathematics
1 answer:
madam [21]1 year ago
6 0

It is given that z varies directly with x and inversely with the square of y so it follows:

z=k\frac{x}{y^2}

It is also given that z=18 when x=6 and y=2 so it follows:

\begin{gathered} 18=k\frac{6}{2^2} \\ k=\frac{18\times4}{6} \\ k=12 \end{gathered}

So the equation of variation becomes:

z=12\frac{x}{y^2}

Therefore the value of z when x=7 and y=7 is given by:

\begin{gathered} z=\frac{12\times7}{7^2} \\ z=\frac{12}{7} \\ z\approx1.7143 \end{gathered}

Hence the value of z is 12/7 or 1.7143.

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I really need help please
anastassius [24]

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Answer:

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slader (a) Find parametric equations for the line through (4, 1, 8) that is perpendicular to the plane x − y + 4z = 2. (Use the
Pavel [41]

Answer:

(x(t), y(t), z(t)) = (4 + t, 1 - t, 8 + 4t)

xy - plane    (x, y, z) = (2, -1, 0)

yz - plane    (x, y, z) = (0, 5, -8)

xz - plane     (x, y, z) = (5, 0, 12)

Step-by-step explanation:

The given point (x, y ,z) = (4, 1, 8)

The plane x -y + 4z = 2

Normal vector (n) = < 1, -1, 4 >

The equation of line through point (4, 1, 8) and the plane is:

(x(t), y(t), z(t)) = (4, 1, 8) + t(1, -1, 4)

(x(t), y(t), z(t)) = (4 + t, 1 - t, 8 + 4t)

Any point on the line P(x, y, z) = ( 4 + t, 1 - t, 8 + 4t)

xy-Pane ⇒ z = 0

8 + 4t = 0

4t = - 8

t = -8/4

t = -2

∴

(x, y, z) = (4 - 2, 1 - 2, 8 + 4(-2))

(x, y, z) = (2, -1, 0)

yz-plane ⇒ x = 0

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1 - t = 0

-t = -1

t = 1

∴

(x, y, z) = ( 4 + 1, 1 - 1, 8 + 4(1) )

(x, y, z) = (5, 0, 12)

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