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3 years ago

Can somebody help with questions 4 plzzzzzx

Mathematics

1 answer:

ivann1987 [24]3 years ago

8 0

Answer:

C, $4.47

Step-by-step explanation:

Firstly you would see that the cost of 5.2 pounds of grapes is $7.75, so you would divide to find the cost of 1 pound of grapes so 7.75/5.2= 1.49 (round to the nearest 100). Then, times 1.49 by 3. Then you get $4.47

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3 years ago

HELP ASAPP PLAZZ :((

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3 years ago

A guitar string is plucked at a distance of 0.6 centimeters above its resting position and then released, causing vibration. The

liubo4ka [24]

Answer:

The equation that represents the motion of the string is given by:

$y =Ae^{-kt}\cos(2\pi ft)$ .....[1] where t represents the time in second.

Given that: A = 0.6 cm (distance above its resting position) , k = 1.8(damping constant) and frequency(f) = 105 cycles per second.

Substitute the given values in [1] we get;

$y =0.6e^{-1.8t}\cos(2\pi 105t)$

$y =0.6e^{-1.8t}\cos(210\pi t)$

(a)

The trigonometric function that models the motion of the string is given by:

$y =0.6e^{-1.8t}\cos(210\pi t)$

(b)

Determine the amount of time t that it takes the string to be damped so that $-0.24\leq y \leq0.24$

Using graphing calculator for the equation

$y =0.6e^{-1.8x}\cos(210\pi x)$

let x = t (time in sec)

Graph as shown below in the attachment:

we get:

the amount of time t that it takes the string to be damped so that $-0.24\leq y \leq0.24$ is, 0.5 sec

4 0

3 years ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .