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3 years ago

1. Seekor burung helang berada 4.56 m di atas aras laut manakala seekor kuda laut berada 6.82 m

Mathematics

1 answer:

s2008m [1.1K]3 years ago

6 0

(a) Calculate the vertical distance between the eagle and the seahorse.

Answer a: 11.38 meters is the vertical distance between them.

(b) Describe Adrian's position from the sea level.

Answer b: Adrian is 1.13m below the sea level.

Step-by-step explanation:

(a) Since sea level is the origin of coordinates. distances above sea level are positive and distances below sea level are negative. The distance between two points is the difference between them.

The vertical distance between the eagle and the seahorse is the difference between their positions:

Distance = 4.56m - (-6.82m) = 4.56m + 6.82m = 11.38 meters

Answer a) 11.38 meters is the vertical distance between them.

(b) Adrian's position is in the middle of the distance between the eagle and the seahorse:

Middle = 11.38m/2 = 5.69m

Since 5.69m is more than 4.56m above the sea level, then Adrian is below the sea level:

4.56m - 5.69m = -1.13m

Answer b: Adrian is 1.13m below the sea level.

<h2><em>Spymore</em></h2>

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Step-by-step explanation:

Note that there are multiple ways to denote a vector. For example, vector u can be written either in bold typeface "u" or with an arrow above it $\vec{u}$ . This explanation uses both representations.

$\displaystyle \vec{u} = \langle 11, 12\rangle =\left(\begin{array}{c}11 \\12\end{array}\right)$ .

$\displaystyle \vec{v} = \langle -16, 6\rangle= \left(\begin{array}{c}-16 \\6\end{array}\right)$ .

$\displaystyle \vec{w} = \langle 4, -5\rangle=\left(\begin{array}{c}4 \\-5\end{array}\right)$ .

There are two components in each of the three vectors. For example, in vector u, the first component is 11 and the second is 12. When multiplying a vector with a constant, multiply each component by the constant. For example,

$3\;\vec{v} = 3\;\left(\begin{array}{c}11 \\12\end{array}\right) = \left(\begin{array}{c}3\times 11 \\3 \times 12\end{array}\right) = \left(\begin{array}{c}33 \\36\end{array}\right)$ .

So is the case when the constant is negative:

$-2\;\vec{v} = (-2)\; \left(\begin{array}{c}-16 \\6\end{array}\right) =\left(\begin{array}{c}(-2) \times (-16) \\(-2)\times(-6)\end{array}\right) = \left(\begin{array}{c}32 \\12\end{array}\right)$ .

When adding two vectors, add the corresponding components (this phrase comes from Wolfram Mathworld) of each vector. In other words, add the number on the same row to each other. For example, when adding 3u to (-2)v,

$3\;\vec{u} + (-2)\;\vec{v} = \left(\begin{array}{c}33 \\36\end{array}\right) + \left(\begin{array}{c}32 \\12\end{array}\right) = \left(\begin{array}{c}33 + 32 \\36+12\end{array}\right) = \left(\begin{array}{c}65\\48\end{array}\right)$ .

Apply the two rules for the four vector operations.

$\displaystyle \begin{aligned}3\;\vec{u} - 2\;\vec{v} + \vec{w} &= 3\;\left(\begin{array}{c}11 \\12\end{array}\right) + (-2)\;\left(\begin{array}{c}-16 \\6\end{array}\right) + \left(\begin{array}{c}4 \\-5\end{array}\right)\\&= \left(\begin{array}{c}3\times 11 + (-2)\times (-16) + 4\\ 3\times 12 + (-2)\times 6 + (-5) \end{array}\right)\\&=\left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle\end{aligned}$

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Now let's make q larger. If w = 12 and q = 20, then w-q = 12-20 = -8 assuming we subtract in the same order. We use absolute value bars to ensure the result is positive. So instead we say

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