All categories

2 years ago

explain why the sine of the acute angle of V is the same regardless of which right triangle is used to find its value

Mathematics

1 answer:

balu736 [363]2 years ago

8 0

Answer:

Because triangle just used to help.

P.S. I’m not sure whether I answer your question

You might be interested in

Pleas help me with my question

sdas [7]

Answer:

Step-by-step explanation:

count the blocks

4 0

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 $t = 6\; \text{hour}$ :

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

7 0

2 years ago

Use the following function rule to find the missing value in the table

gladu [14]

The answer is c) -14.8

6 0

3 years ago

The length l, width w, and height h of a box change with time. At a certain instant the dimensions are l = 3 m and w = h = 6 m,

Gemiola [76]

Answer:

a) The rate of change associated with the volume of the box is 54 cubic meters per second, b) The rate of change associated with the surface area of the box is 18 square meters per second, c) The rate of change of the length of the diagonal is -1 meters per second.

Step-by-step explanation:

a) Given that box is a parallelepiped, the volume of the parallelepiped, measured in cubic meters, is represented by this formula:

$V = w \cdot h \cdot l$

Where:

$w$ - Width, measured in meters.

$h$ - Height, measured in meters.

$l$ - Length, measured in meters.

The rate of change in the volume of the box, measured in cubic meters per second, is deducted by deriving the volume function in terms of time:

$\dot V = h\cdot l \cdot \dot w + w\cdot l \cdot \dot h + w\cdot h \cdot \dot l$

Where $\dot w$ , $\dot h$ and $\dot l$ are the rates of change related to the width, height and length, measured in meters per second.

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the volume of the box is:

$\dot V = (6\,m)\cdot (3\,m)\cdot \left(3\,\frac{m}{s} \right)+(6\,m)\cdot (3\,m)\cdot \left(-6\,\frac{m}{s} \right)+(6\,m)\cdot (6\,m)\cdot \left(3\,\frac{m}{s}\right)$

$\dot V = 54\,\frac{m^{3}}{s}$

The rate of change associated with the volume of the box is 54 cubic meters per second.

b) The surface area of the parallelepiped, measured in square meters, is represented by this model:

$A_{s} = 2\cdot (w\cdot l + l\cdot h + w\cdot h)$

The rate of change in the surface area of the box, measured in square meters per second, is deducted by deriving the surface area function in terms of time:

$\dot A_{s} = 2\cdot (l+h)\cdot \dot w + 2\cdot (w+h)\cdot \dot l + 2\cdot (w+l)\cdot \dot h$

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the surface area of the box is:

$\dot A_{s} = 2\cdot (6\,m + 3\,m)\cdot \left(3\,\frac{m}{s} \right) + 2\cdot (6\,m+6\,m)\cdot \left(3\,\frac{m}{s} \right) + 2\cdot (6\,m + 3\,m)\cdot \left(-6\,\frac{m}{s} \right)$

$\dot A_{s} = 18\,\frac{m^{2}}{s}$

The rate of change associated with the surface area of the box is 18 square meters per second.

c) The length of the diagonal, measured in meters, is represented by the following Pythagorean identity:

$r^{2} = w^{2}+h^{2}+l^{2}$

The rate of change in the surface area of the box, measured in square meters per second, is deducted by deriving the surface area function in terms of time before simplification:

$2\cdot r \cdot \dot r = 2\cdot w \cdot \dot w + 2\cdot h \cdot \dot h + 2\cdot l \cdot \dot l$

$r\cdot \dot r = w\cdot \dot w + h\cdot \dot h + l\cdot \dot l$

$\dot r = \frac{w\cdot \dot w + h \cdot \dot h + l \cdot \dot l}{\sqrt{w^{2}+h^{2}+l^{2}}}$

Given that $w = 6\,m$ , $h = 6\,m$ , $l = 3\,m$ , $\dot w =3\,\frac{m}{s}$ , $\dot h = -6\,\frac{m}{s}$ and $\dot l = 3\,\frac{m}{s}$ , the rate of change in the length of the diagonal of the box is:

$\dot r = \frac{(6\,m)\cdot \left(3\,\frac{m}{s} \right)+(6\,m)\cdot \left(-6\,\frac{m}{s} \right)+(3\,m)\cdot \left(3\,\frac{m}{s} \right)}{\sqrt{(6\,m)^{2}+(6\,m)^{2}+(3\,m)^{2}}}$

$\dot r = -1\,\frac{m}{s}$

The rate of change of the length of the diagonal is -1 meters per second.

6 0

3 years ago

I don’t know how to answer this please help

LUCKY_DIMON [66]

Answer:

I think 38%

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers