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3 years ago

Value of the derivative of g(x)=8-10Cosx at 'x=0' is?

Mathematics

1 answer:

VLD [36.1K]3 years ago

3 0

Answer:

g'(0) = 0

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Algebra I</u>

Function Notation

<u>Pre-Calculus</u>

Unit Circle

<u>Calculus</u>

Derivatives
Derivative Notation
The derivative of a constant is equal to 0
Derivative Property: $\frac{d}{dx} [cf(x)] = c \cdot f'(x)$
Trig Derivative: $\frac{d}{dx} [cos(x)] = -sin(x)$

Step-by-step explanation:

<u>Step 1: Define</u>

g(x) = 8 - 10cos(x)

x = 0

<u>Step 2: Differentiate</u>

Differentiate [Trig]: g'(x) = 0 - 10[-sin(x)]
Simplify Derivative: g'(x) = 10sin(x)

<u>Step 3: Evaluate</u>

Substitute in <em>x</em>: g'(0) = 10sin(0)
Evaluate Trig: g'(0) = 10(0)
Multiply: g'(0) = 0

You might be interested in

Find the angle between u =the square root of 5i-8j and v =the square root of 5i+j.

fenix001 [56]

Answer:

The angle between vector $\vec{u} = 5\, \vec{i} - 8\, \vec{j}$ and $\vec{v} = 5\, \vec{i} + \, \vec{j}$ is approximately $1.21$ radians, which is equivalent to approximately $69.3^\circ$ .

Step-by-step explanation:

The angle between two vectors can be found from the ratio between:

their dot products, and
the product of their lengths.

To be precise, if $\theta$ denotes the angle between $\vec{u}$ and $\vec{v}$ (assume that $0^\circ \le \theta < 180^\circ$ or equivalently $0 \le \theta < \pi$ ,) then:

$\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ .

<h3>Dot product of the two vectors</h3>

The first component of $\vec{u}$ is $5$ and the first component of $\vec{v}$ is also

The second component of $\vec{u}$ is $(-8)$ while the second component of $\vec{v}$ is $1$ . The product of these two second components is $(-8) \times 1= (-8)$ .

The dot product of $\vec{u}$ and $\vec{v}$ will thus be:

$\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}$ .

<h3>Lengths of the two vectors</h3>

Apply the Pythagorean Theorem to both $\vec{u}$ and $\vec{v}$ :

$\| u \| = \sqrt{5^2 + (-8)^2} = \sqrt{89}$ .
$\| v \| = \sqrt{5^2 + 1^2} = \sqrt{26}$ .

<h3>Angle between the two vectors</h3>

Let $\theta$ represent the angle between $\vec{u}$ and $\vec{v}$ . Apply the formula $\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ to find the cosine of this angle:

$\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}$ .

Since $\theta$ is the angle between two vectors, its value should be between $0\; \rm radians$ and $\pi \; \rm radians$ ( $0^\circ$ and $180^\circ$ .) That is: $0 \le \theta < \pi$ and $0^\circ \le \theta < 180^\circ$ . Apply the arccosine function (the inverse of the cosine function) to find the value of $\theta$ :

$\displaystyle \cos^{-1}\left(\frac{17}{\sqrt{89}\cdot \sqrt{26}}\right) \approx 1.21 \;\rm radians \approx 69.3^\circ$ .

3 0

3 years ago

Find the distance between (-3, 11) and (-6,4).Round to the nearest then the

Likurg_2 [28]

Answer:

d =7.6

Step-by-step explanation:

$(-3,11) =(x_1,y_1)\\(-6,4)=(x_2,y_2)\\\\\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\\\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\\\\=\sqrt{\left(-6-\left(-3\right)\right)^2+\left(4-11\right)^2}\\= \sqrt{(-6+3)^2+(4-11)^2}\\ \\d= \sqrt{(-3)^2+(-7)^2} \\\\d = \sqrt{9+49}\\ \\d = \sqrt{58}\\\\d = 7.61577\\\\d =7.6$