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OLEGan [10]
3 years ago
11

2.6 cm = .........................mm

Mathematics
1 answer:
just olya [345]3 years ago
8 0

Answer:

26 mm

Step-by-step explanation:

1 cm is equal to 10 mm

2.6 cm

> 2.6 x 10

> 26 mm

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Determine the mean of 28,40,53,39,45
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Let f be the function defined by f(x) = e^(x) cos x.
Pavel [41]
(a)

The average rate of change of f on the interval 0 ≤ x ≤ π is

   \displaystyle
f_{avg\Delta} = \frac{f(\pi) - f(0)}{\pi - 0} =\frac{-e^\pi-1}{\pi}

____________

(b)

f(x) = e^{x} cos x \implies f'(x) = e^x \cos(x) - e^x \sin(x) \implies \\ \\
f'\left(\frac{3\pi}{2} \right) = e^{3\pi/2} \cos(3\pi/2) - e^{3\pi/2} \sin(3\pi/2) \\ \\
f'\left(\frac{3\pi}{2} \right) = 0 - e^{3\pi/2} (-1) = e^{3\pi/2}

The slope of the tangent line is e^{3\pi/2}.

____________

(c)

The absolute minimum value of f occurs at a critical point where f'(x) = 0 or at endpoints.

Solving f'(x) = 0

f'(x) = e^x \cos(x) - e^x \sin(x) \\ \\
0 = e^x \big( \cos(x) - \sin(x)\big)

Use zero factor property to solve.

e^x \ \textgreater \  0\forall x \in \mathbb{R} so that factor will not generate solutions.
Set cos(x) - sin(x) = 0

\cos (x) - \sin (x) = 0 \\
\cos(x) = \sin(x)

cos(x) = 0 when x = π/2, 3π/2, but x = π/2. 3π/2 are not solutions to the equation. Therefore, we are justified in dividing both sides by cos(x) to make tan(x):

\displaystyle\cos(x) = \sin(x) \implies 0 = \frac{\sin (x)}{\cos(x)} \implies 0 = \tan(x) \implies \\ \\
x = \pi/4,\ 5\pi/4\ \forall\ x \in [0, 2\pi]

We check the values of f at the end points and these two critical numbers.

f(0) = e^1 \cos(0) = 1

\displaystyle f(\pi/4) = e^{\pi/4} \cos(\pi/4) = e^{\pi/4}  \frac{\sqrt{2}}{2}

\displaystyle f(5\pi/4) = e^{5\pi/4} \cos(5\pi/4) = e^{5\pi/4}  \frac{-\sqrt{2}}{2} = -e^{\pi/4}  \frac{\sqrt{2}}{2}

f(2\pi) = e^{2\pi} \cos(2\pi) = e^{2\pi}

There is only one negative number.
The absolute minimum value of f <span>on the interval 0 ≤ x ≤ 2π is
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____________

(d)

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g is a differentiable function; therefore, it is a continuous function, which tells us \lim_{x \to \pi/2} g(x) = g(\pi/2) = 0.

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f'(\pi/2) = e^{\pi/2} \big( \cos(\pi/2) - \sin(\pi/2)\big) = -e^{\pi/2} \\ \\&#10;g'(\pi/2) = 2

thus

\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \frac{-e^{\pi/2}}{2}</span>

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