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3 years ago

Question 7 of 10

Mathematics

2 answers:

Vsevolod [243]3 years ago

6 0

Answer:

g(x) = –3(x + 6)2 + 48.

Step-by-step explanation:

The graph of f(x) = x2 is shifted right 6 units. The graph of f(x) = x2 is shifted down 48 units. The graph of f(x) = x2 is reflected over the y-axis.

Mariulka [41]3 years ago

6 0

Answer:

Answer for this equation: h(x) = 3x2 - 7?

The graph of f(x) is vertically stretched by a factor of 3 and shifted

7 units down.

Step-by-step explanation:

Ap3x Confirmed

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user100 [1]

Answer:

1/64

Step-by-step explanation:

1/64=(1/8)^2

1/64=(1/4)^3

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3 years ago

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Flying fish use their pectoral fins like airplane wings to glide through the air. Suppose a flying fish reaches a maximum height

GrogVix [38]

The flight is in the shape of a parabola with a vertex 5 feet above the water and 1/2 * 33 = 16.5 feet horizontally from the point of leaving the water

y = a(x - h)^2 + k

where (h,k) is the vertex of the parabola and here it is (5 , 16.5), so we have the function:-

y = a(x - 16.5)^2 + 5

when x = 0 y = 0 so

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which gives a = -0.018365

So our function for the flight path is

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3 years ago

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morpeh [17]

Answer:

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3 years ago

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-

ivanzaharov [21]

Answer:

$c = \displaystyle\frac{9}{4}$

Step-by-step explanation:

The following information is missing in the given question:

$f(x) = \sqrt{x}$

Using this we may solve the question as:

We are given the following in the question:

$f(x) = \sqrt{x}, x \in [0,9]$

We have to find the number c such that f(x) satisfies the Mean value theorem.

Mean Value theorem:

It states that if the function is differentiable in the closed interval [a,b], differentiable in the interval (a,b), then there exist c in (a,b) such that:

$f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}$

Now,

$f(x) = \sqrt{x}\\f'(x) = \frac{1}{2\sqrt{x}}$

Continuity in [0,9]

Since a polynomial function is continuous everywhere, f(x) is continuous in [0,9]

Differentiability in (0,9)

Since a polynomial function is differentiable everywhere the given function is differentiable in interval (0,9)

Then, by mean value theorem:

$f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}\\\\\frac{1}{2\sqrt{c}} = \frac{f(9) - f(0)}{9-0} = \frac{3}{9}\\\\\frac{1}{2\sqrt{c}} = \frac{1}{3}\\\\c= \frac{9}{4}$

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Dovator [93]

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