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3 years ago

5

Mat told Sam that for a relation to be a function the input values can repeat but the output values cannot. Sam thought Mrs. Gul

likson told him in
class that if the output values repeat and have different input values, then the relation is not a function. Explain who is correct.
In you explanation tell who is correct, explain how you know (hint make sure you explain the definition of a function in your answer), and include
an example of a function that they could show to prove they are right. Hint Look at lesson 1.1

1 answer:

Masteriza [31]3 years ago

4 0

Answer:

Step-by-step explanation:

a

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28 + 56t + 28w equivalent expressions

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3 years ago

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 $\theta$ :

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

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