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Dovator
3 months ago
9

Jackson is conducting an experiment for his Physics class. He attaches a weight to the bottom of a metal spring. He then pulls t

he weight down so that it is a distance of six inches from its equilibrium position. Jackson then releases the weight and finds that it takes four seconds for the spring to complete one oscillation. Which of the following functions best models the position of the weight?
Mathematics
2 answers:
zzz [12.3K]3 months ago
7 0
The behavior of the spring can be described using either a sine or cosine function. The spring's maximum displacement is 6 inches, occurring at t=0, which we will define as the positive peak. Therefore, we can express the function as:
6sin(at+B). The spring's period is 4 minutes, which means the time factor in the equation must complete a cycle (2π) in 4 minutes. This gives us the equation 4min*a=2π, leading to a=π/2. Generally, a=2π/T where a is the coefficient and T is the period. For B, since sin(π/2)=1, we determine B=π/2 because at t=0, the equation becomes 6sin(B)=6. Therefore, we substitute to form f(t)=6sin(πt/2+π/2)=6cos(πt/2)
due to trigonometric relations.
babunello [11.8K]3 months ago
5 0

Answer:

The correct choice for plato is C

s(t) = 6cos (π/2*t)

Step-by-step explanation:


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The accurate statements are 0.75 (x,y) = (0.75x, 0.75y), LM is parallel to L'M', and "the vertices of the image are nearer to the origin relative to the pre-image".

Explanation:

It is stated that the triangle ABC underwent dilation as per the rule DO,0.75 (x,y)

DO, 0.75(x,y) signifies the dilation rule, where both x and y values are scaled by 0.75, indicating that the dilation, centered around the origin, has a scale factor of 0.75. The representation for this dilation would be,

(x,y)\rightarrow (0.75x,0.75y)

Consequently, the first statement holds true.

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Thus, the second statement "LM is parallel to L'M'" is also accurate, while the third and fifth statements are incorrect.

The scale factor of 0.75, being less than 1, suggests that the vertices of the image are closer to the origin compared to those of the pre-image.

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\bf \qquad \qquad \textit{Annual Yield Formula}
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\left(1+\frac{0.040784}{12}\right)^{12}-1\\\\
-------------------------------\\\\
~~~~~~~~~~~~\textit{4.0798\% compounded semiannually}\\\\


\bf ~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
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\begin{cases}
r=rate\to 4.0798\%\to \frac{4.0798}{100}\to &0.040798\\
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\left(1+\frac{0.040798}{2}\right)^{2}-1\\\\
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\bf ~~~~~~~~~~~~\textit{4.0730\% compounded daily}\\\\
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