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3 years ago

Please help me with the domain and range this is Algebra 2

Mathematics

1 answer:

adelina 88 [10]3 years ago

3 0

Domain: The set of all real numbers
We can plug in any x value we want and there aren't any restrictions. There aren't any holes, jumps, gaps, or asymptotes

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Range: Only one value is in the range and it is the value 2. We can write this as {2}. The curly brackets indicate "set". It is possible to have a set of only 1 item inside.

The range is the set of y value outputs. In this case, the output is 2 because we have a horizontal line. No matter what the x input is, the output is always 2. This is known as a constant function. The output is constantly the same.

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3 years ago

Read 2 more answers

I need help with this please its due today :(

yKpoI14uk [10]

Answer:

$a) \ 6^2 \times 6^3 = 6^{2+3} = 6^5 = 7776\\\\b)\ \frac{(-3)^5 }{(-3)^2} = (-3)^{5-2} = (-3)^3 = -27\\\\c)\ [(-2)^3]^4 = (-2)^{3\times 4} = (-2)^{12} = 4096 \\\\d)\ (\frac{1}{999})^0 = 1, anything \ raise \ to \ zero = 1\\\\e) \ (\frac{4}{3})^{-2} = 4^{-2} \times 3^2 = \frac{9}{16}\\\\f) \ \frac{2^5 \times 2^9}{2^{16}} = 2^{5+9-16} = 2^{-2} = \frac{1}{4}$

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3 years ago

A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

ira [324]

Answer:

Approximately $58.2\; \text{nautical miles}$ (assuming that the bearing is ${\rm S$29^{\circ}$W}$ .)

Step-by-step explanation:

Let $v$ denote the speed of the ship, and let $t$ denote the duration of the trip. The magnitude of the displacement of this ship would be $v\, t$ .

Refer to the diagram attached. The direction ${\rm S$29^{\circ}$W}$ means $29^{\circ}$ west of south. Thus, start with the south direction and turn towards west (clockwise) by $29^{\circ}$ to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of $v\, t$ . The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly $29^{\circ}$ .

Since the hypotenuse is of length $v\, t$ , the dashed line segment opposite to the $\theta = 29^{\circ}$ vertex would have a length of:

$\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}$ .

Substitute in $\begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned}$ and $t = 6\; \text{hour}$ :

$\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}$ .

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2 years ago