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marusya05 [52]
3 years ago
6

H(x) = 11 - |4 - 4x| find h(x) = 4

Mathematics
1 answer:
liubo4ka [24]3 years ago
3 0
Plug in 4 as x
H(x) =11 - | 4 - 4(4)|
= 11 - |4 - 16|
Subtract 4 and 16
= 11-|-12|
The negative goes away because of the absolute value sign
11- | 12|
Then, subtract
H(x) = -1
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Find the width of the rectangular prism when the surface area is 208 square centimeters.
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SA=2(lw+wh+lh) This is the formula for finding the surface area of a rectangular prism, where SA is surface area, l is length, w is width, and h is height.

208=2(lw+wh+lh)
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4 years ago
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3 years ago
slader (a) Find parametric equations for the line through (4, 1, 8) that is perpendicular to the plane x − y + 4z = 2. (Use the
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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

4 + t = 0

t = -4

∴

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

(x, y, z) = (0, 5, -8)

xz-plane ⇒ y = 0

1 - t = 0

-t = -1

t = 1

∴

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

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

6 0
3 years ago
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