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3 years ago

evaluate the function h(x) = |-2x|-9 for the given values of x a) h(4)=. b) h(-5)=. c) h(0) =

Mathematics

1 answer:

Nikitich [7]3 years ago

6 0

Answer:

h(4) = -1 ; h(-5) = 1 h(0) = -9

Step-by-step explanation:

a) h(x) = |-2x| - 9 Substitute x = 4

h(4) = | -2 · 4| - 9

h(x) = | -8| - 9

h(4) = 8 - 9

h(4) = -1

b) h(x) = |-2x| - 9 Substitute x = -5

h(-5) = |-2 · -5| - 9

h(-5) = |-10| - 9

h(-5) = 10 - 9

h(-5) = 1

c) h(x) = |-2x| - 9 Substitute x = 0

h(0) = |-2 · 0| - 9

h(0) = |0| - 9

h(0) = 0 - 9

h(0) = -9

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3 years ago

A giant tank in a shape of an inverted cone is filled with oil. the height of the tank is 1.5 metre and its radius is 1 metre. t

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The given height of the cylinder of 1.5 m, and radius of 1 m, and the rate

of dripping of 110 cm³/s gives the following values.

1) The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ 9.34 × 10⁻⁵ m/s

2) The rate of change of the oil's height when the height is 20 cm is h' ≈ 1.97 × 10⁻³ m/s

3) The rate the oil radius is changing when the radius is 10 cm is approximately 0.175 m/s

<h3>How can the rate of change of the radius & height be found?</h3>

The given parameters are;

Height of the tank, h = 1.5 m

Radius of the tank, r = 1 m

Rate at which the oil is dripping from the tank = 110 cm³/s = 0.00011 m³/s

$1) \hspace{0.15 cm}V = \frac{1}{3} \cdot \pi \cdot r^2 \cdot h$

From the shape of the tank, we have;

$\dfrac{h}{r} = \dfrac{1.5}{1}$

Which gives;

h = 1.5·r

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)}$

$\dfrac{d}{dr} V =\dfrac{d}{dr} \left( \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)\right) = \dfrac{3}{2} \cdot \pi \cdot r^2$

$\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dr} }}$

$\dfrac{dV}{dt} = 0.00011$

Which gives;

$\dfrac{dr}{dt} = \mathbf{ \dfrac{0.00011 }{\dfrac{3}{2} \cdot \pi \cdot r^2}}$

When r = 0.5 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.5^2} \approx 9.34 \times 10^{-5}$

The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ 9.34 × 10⁻⁵ m/s

2) When the height is 20 cm, we have;

h = 1.5·r

$r = \dfrac{h}{1.5}$

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot \left(\dfrac{h}{1.5} \right) ^2 \cdot h}$

r = 20 cm ÷ 1.5 = $13.\overline3$ cm = $0.1\overline3$ m

Which gives;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.1 \overline{3}^2} \approx \mathbf{1.313 \times 10^{-3}}$

$\dfrac{d}{dh} V = \dfrac{d}{dh} \left(\dfrac{4}{27} \cdot \pi \cdot h^3 \right) = \dfrac{4 \cdot \pi \cdot h^2}{9}$

$\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dh} }$

$\dfrac{dh}{dt} = \mathbf{\dfrac{0.00011}{\dfrac{4 \cdot \pi \cdot h^2}{9}}}$

When the height is 20 cm = 0.2 m, we have;

$\dfrac{dh}{dt} = \dfrac{0.00011}{\dfrac{4 \times \pi \times 0.2^2}{9}} \approx \mathbf{1.97 \times 10^{-3}}$

The rate of change of the oil's height when the height is 20 cm is h' ≈ 1.97 × 10⁻³ m/s

3) The volume of the slick, V = π·r²·h

Where;

h = The height of the slick = 0.1 cm = 0.001 m

Therefore;

V = 0.001·π·r²

$\dfrac{dV}{dr} = \mathbf{ 0.002 \cdot \pi \cdot r}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{0.00011 }{0.002 \cdot \pi \cdot r}}$

When the radius is 10 cm = 0.1 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{0.002 \times \pi \times 0.1} \approx \mathbf{0.175}$

The rate the oil radius is changing when the radius is 10 cm is approximately 0.175 m

Learn more about the rules of differentiation here:

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3 years ago

What are the apparent zeros of the cubic function graphed above? (also if you have the answers to the rest of this test plz send

Mazyrski [523]

The point where a function (for our problem its the cubic function shown)

cuts/crosses the x-axis

and/or

touches the x-axis

is a zero of the function.

From the graph shown, we can clearly see that it cuts the x-axis at -1 and touches the x-axis at 2.

So the zeros are at -1 and 2.

ANSWER: {-1,2}

7 0

4 years ago

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klasskru [66]

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3 years ago

If F(x) = +4, which of the following is the inverse of Fx)?

Viefleur [7K]

Answer:

$f^{-1} = \frac{ 7(x-4)}{2}$

Step-by-step explanation:

Given

$f(x) = \frac{2x}{7} + 4$

Required

The inverse function

$f(x) = \frac{2x}{7} + 4$

Replace f(x) with y

$y = \frac{2x}{7} + 4$

Swap x and y

$x = \frac{2y}{7} + 4$

Multiply through by 7

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Solve for 2y

$2y = 7x - 28$

Solve for y

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Factorize

$y = \frac{ 7(x-4)}{2}$

Write as:

$f^{-1} = \frac{ 7(x-4)}{2}$

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3 years ago