Least to greatest .5 1/5 .35 12/25 4/5

Towns P,Q,R and S are shown. Q is 35 km due East of P S is 15km due West of P R is 15km due South of P Work out the bearing of R

Anni [7]

Answer:

Part A

The bearing of the point 'R' from 'S' is 225°

Part B

The bearing from R to Q is approximately 293.2°

Step-by-step explanation:

The location of the point 'Q' = 35 km due East of P

The location of the point 'S' = 15 km due West of P

The location of the 'R' = 15 km due south of 'P'

Part A

To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that $\overline{RP}$ = $\overline{SP}$ , the right triangle ΔRPS is an isosceles right triangle

∴ ∠PRS = ∠PSR = 45°

The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°

Part B

∠PRQ = arctan(35/15) ≈ 66.8°

Therefore the bearing from R to Q = 270 + 90 - 66.8 ≈ 293.2°

6 0

3 years ago

Exponential Growth and Decay The expressions below represent either exponential growth or exponential decay. Check all the expre

Tomtit [17]

Answer:

First one: 5x3(0.75)2t

and fourth one: 0.05x500(0.005)t

Step-by-step explanation:

In order to have an exponencial decay, the rate of the exponencial function needs to be lesser than 1.

The rate is the value between parenthesis, so in the first equation, the rate is 0.75, so this is an exponencial decay.

The second equation has rate = 2.4, so this is not an exponencial decay.

The third equation has rate = 1.04, so this is not an exponencial decay.

The fourth equation has rate = 0.005, so this is an exponencial decay.

5 0

3 years ago

Read 2 more answers

Graphing Exponential Functions In Exercise, sketch the graph of the function.

denis23 [38]

Answer:

The graph of the function $f(x) = (1/2)^2x + 4$ is attached as image below.

Analysis:

We have two intersection points in the graph

The function $f(x) = (1/2)^2x + 4$ intersects the y-axis at points (0,4)

The function $f(x) = (1/2)^2x + 4$ intersects the x-axis at points (-16,0)

f(x) is positive for x>-`16

f(x) is negative for x<-`16

5 0

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 $\theta$ :