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crimeas [40]
3 years ago
11

$239 television 10% discount

Mathematics
1 answer:
kiruha [24]3 years ago
5 0
23.9 is the discount so you gonna round it up so get an whole number the answer is 24
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A pyramid has a regular hexagonal base with side links of four and a slant height of six. Find the total area of the pyramid
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Area of the Base = 6 * side^2 / 4 * tan (180/6)

Area of the Base = 6 * 16 / 4 * 0.57735

Area of the Base = 96 / 2.3094

Area of the Base = 41.5692387633


Area of 1 Face = 2 * 6 = 12

Area of 6 Faces = 72


Total Area = 41.5692387633 + 72

Total Area = 113.5692387633

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Step-by-step explanation:

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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
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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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