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Marizza181 [45]
1 year ago
12

On a coordinate plane, 2 exponential functions are shown. f (x) decreases from quadrant 2 to quadrant 1 and approaches y = 0. It

crosses the y-axis at (0, 4) and goes through (1, 2). g (x) increases from quadrant 3 into quadrant 4 and approaches y = 0. It crosses the y-axis at (0, negative 4) and goes through (1, negative 2).
Which function represents g(x), a reflection of f(x) = 4(one-half) Superscript x across the x-axis?

g(x) = −4(2)x
g(x) = 4(2)−x
g(x) = −4(one-half) Superscript x
g(x) = 4(one-half) Superscript negative x
Mathematics
1 answer:
aliina [53]1 year ago
3 0

Answer: -4\left(\frac{1}{2} \right)^{x}

Step-by-step explanation:

To reflect the graph of a function across the x-axis,

y=f(x) \longrightarrow y=-f(x)

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identity property

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At what point does she lose contact with the snowball and fly off at a tangent? That is
postnew [5]

Answer:

α ≥ 48.2°

Step-by-step explanation:

The complete question is given as follows:

" A skier starts at the top of a very large frictionless snowball, with a very small initial speed, and skis straight  down the side. At what point does she lose contact with the snowball and fly off at a tangent? That is, at the  instant she loses contact with the snowball, what angle α does a radial line from the center of the snowball to  the skier make with the vertical?"

- The figure is also attached.

Solution:

- The skier has a mass (m) and the snowball’s radius (r).

- Choose the center of the snowball to be the zero of gravitational  potential. - We can look at the velocity (v) as a function of the angle (α) and find the specific α at which the skier lifts off and  departs from the snowball.

- If we ignore snow-­ski friction along with air resistance, then the one work producing force in this problem, gravity,  is conservative. Therefore the skier’s total mechanical energy at any angle α is the same as her total mechanical  energy at the top of the snowball.

- Hence, From conservation of energy we have:

                       KE (α) + PE(α) = KE(α = 0) + PE(α = 0)

                       0.2*m*v(α)^2 + m*g*r*cos(α) = 0.5*m*[ v(α = 0)]^2 + m*g*r

                       0.2*m*v(α)^2 + m*g*r*cos(α) ≈ m*g*r

                        m*v(α)^2 / r = 2*m*g( 1 - cos(α) )

- The centripetal force (due to gravity) will be mgcosα, so the skier will remain on the snowball as long as gravity  can hold her to that path, i.e. as long as:

                         m*g*cos(α) ≥ 2*m*g( 1 - cos(α) )

- Any radial gravitational force beyond what is necessary for the circular motion will be balanced by the normal  force—or else the skier will sink into the snowball.

- The expression for α_lift becomes:

                            3*cos(α) ≥ 2

                            α ≥ arc cos ( 2/3) ≥ 48.2°

4 0
2 years ago
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6 0
3 years ago
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What happens to the mean of the data set {10, 2, 8, 9, 5, 2, 6} if the number 12 is added to the data set? A. The mean increases
Setler [38]

Answer:

C. The mean increases by 0.75.

Step-by-step explanation:

The mean of a data set is given by the sum of all its values, divided by the number of values.

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7 values

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Adding number 12 to the data set

Now we have 8 values

The sum is 42 + 12 = 54

The mean is 54/8 = 6.75

This means that the mean increases by 0.75.

7 0
2 years ago
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