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erastova [34]
2 years ago
12

Write a rule for g described by the transformation of the graph of f...

Mathematics
1 answer:
Svetlanka [38]2 years ago
4 0

Answer:

g(x) = 4-4\sqrt{x}

Step-by-step explanation:

Given

f(x) = 4\sqrt{x}+1

1. 5 units down translation

2. Reflection in the x-axis

Required

Determine the resulting function g

1. 5 units down translation

This is defined by the rule:

A down translation is:

f'(x) = f(x) - b

Where b is the number of units.

So, we have:

f'(x) = 4\sqrt{x} + 1 - 5

f'(x) = 4\sqrt{x} -4

2. Reflection in the x-axis

For a function f(x),

When reflected over the x-axis, it becomes -f(x)

So:

g(x) = -f(x)

g(x) = -(4\sqrt{x} - 4)

g(x) = -4\sqrt{x} +4

g(x) = 4-4\sqrt{x}

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Consider the following equations. f(x) = − 4/ x3, y = 0, x = −2, x = −1. Sketch the region bounded by the graphs of the equation
Sindrei [870]

Answer:

1.5 unit^2

Step-by-step explanation:

Solution:-

- A graphing utility was used to plot the following equations:

                         f ( x ) = - \frac{4}{x^3}\\\\y = 0 , x = -1 , x = -2

- The plot is given in the document attached.

- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).

- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.

We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).

The double integration formulation can be written as:

                           A= \int\limits_c^d \int\limits_a^b {} \, dy.dx \\\\A = \int\limits_c^d { - \frac{4}{x^3} } . dx\\\\A = \frac{2}{x^2} |\limits_-_2^-^1\\\\A = \frac{2}{1} - \frac{2}{4} \\\\A = \frac{3}{2} unit^2

Answer: 1.5 unit^2 is the amount of area bounded by the given curve f ( x ) and the boundary equations.

Download docx
3 0
2 years ago
The diameter of a circle has endpoints P(-12, -4) and Q(6, 12).
nexus9112 [7]

9514 1404 393

Answer:

  (x +3)² +(y -4)² = 145

Step-by-step explanation:

The center of the circle is the midpoint of the given segment PQ. If we call that point A, then ...

  A = (P +Q)/2

  A = ((-12, -4) +(6, 12))/2 = (-12+6, -4+12)/2 = (-6, 8)/2

  A = (-3, 4)

The equation of the circle for some radius r is ...

  (x -(-3))² +(y -4)² = r² . . . . . . where (-3, 4) is the center of the circle

The value of r² can be found by substituting either of the points on the circle. If we use Q, then we have ...

  (6 +3)² +(12 -4)² = r² = 9² +8²

  r² = 81 +64 = 145

Then the equation of the circle is ...

  (x +3)² +(y -4)² = 145

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3 years ago
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The answer to this is:

x>24

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Step-by-step explanation:

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this may be right but I do not know

7 0
2 years ago
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