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Yakvenalex [24]
1 year ago
14

Please help!! I tried solving this, but it keeps saying I'm wrong. I'm genuinely confused on how to do this...

Mathematics
1 answer:
Wewaii [24]1 year ago
5 0

We are given a general cosine function, that is:

y=A\cos (B(x-C))+D

Where:

A=\text{ amplitude}

The amplitude "A" is half the distance between the maximum and minimum values of the function, in this case the maximum value is 3.7 and the minimum value is 0.3, therefore, the amplitude is:

A=\frac{3.7-0.3}{2}=1.7

Therefore, A = 1.7

B=frequency

The value "B" is the number of cycles it takes the function to go from 0 to 2pi

The number B is usually expressed as:

T=\frac{2\pi}{B}

"T" is called the period and is how much it takes the curve to do one cycle, in this case the period is:

T=3.9-2.3=1.6

Therefore, replacing we get:

\begin{gathered} 1.6=\frac{2\pi}{B} \\ B=\frac{2\pi}{1.6} \end{gathered}

The number C is called the phase and refers to how much is the function is shifted horizontally, in this case, the function is shifted by 2.3 in the positive direction, therefore C = 2.3.

The number D is how much the graph is shifted vertically, in this case, it is shifted by 0.3 plus the amplitude, that is:

D=0.3+1.7=2

Replacing the values we get:

y=1.7\cos (\frac{2\pi}{1.6}(x-2.3))+2

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The discriminant is b^2-4ac
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11^2-4(1)(-10)
1. Simplify 11^2 to 121
121-4(1)(-10)

2. Simplify 4*1*-10 to -40
121-(-40)

3. Simplify brackets
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4. Simplify 
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Your discriminant is 161


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What is 10 8by13 add 4 24by49. these numbers are in mixed fraction​
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Step-by-step explanation:

10 8/13 + 4 24/49

138/13 + 220/49

taking LCM

(138*49 + 220*13) / (13*49)

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Step-by-step explanation:

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3 years ago
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A rock thrown vertically upward from the surface of the moon at a velocity of 36​m/sec reaches a height of s = 36t - 0.8 t^2 met
Verdich [7]

Answer:

a. The rock's velocity is v(t)=36-1.6t \:{(m/s)}  and the acceleration is a(t)=-1.6  \:{(m/s^2)}

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

  • Velocity is defined as the rate of change of position or the rate of displacement. v(t)=\frac{ds}{dt}
  • Acceleration is defined as the rate of change of velocity. a(t)=\frac{dv}{dt}

a.

The rock's velocity is the derivative of the height function s(t) = 36t - 0.8 t^2

v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t

The rock's acceleration is the derivative of the velocity function v(t)=36-1.6t

a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6

b. The rock will reach its highest point when the velocity becomes zero.

v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = \frac{405}{2} =202.5 \:m

To find the time it reach half its maximum height, we need to solve

36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45

The rock is aloft for 45 seconds.

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