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The light from the moon, in lux, on the night of the t^("th ") day of 2016 , is L(t)=0.25-sin((2pi(t-2))/(28.5)) What is the period of the light from the moon? Give an exact answer. days

The light from the moon, in lux, on the night of the ttht^{\text{th}} day of 20162016 , is\newlineL(t)=0.25−sin⁡(2π(t−2)28.5)L(t)=0.25-\sin\left(\frac{2\pi(t-2)}{28.5}\right)\newlineWhat is the period of the light from the moon? Give an exact answer.\newline days

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Q. The light from the moon, in lux, on the night of the ttht^{\text{th}} day of 20162016 , is\newlineL(t)=0.25−sin⁡(2π(t−2)28.5)L(t)=0.25-\sin\left(\frac{2\pi(t-2)}{28.5}\right)\newlineWhat is the period of the light from the moon? Give an exact answer.\newline days
  1. Identify General Form: Identify the general form of a sinusoidal function.\newlineA general sinusoidal function can be written as f(t)=A⋅sin⁡(B(t−C))+Df(t) = A \cdot \sin(B(t - C)) + D or f(t)=A⋅cos⁡(B(t−C))+Df(t) = A \cdot \cos(B(t - C)) + D, where:\newline- AA is the amplitude,\newline- BB is related to the period (PP) of the function by the formula P=2πBP = \frac{2\pi}{B},\newline- CC is the phase shift, and\newline- DD is the vertical shift.
  2. Compare Given Function: Compare the given function to the general form.\newlineThe given function is L(t)=0.25−sin⁡(2π(t−2)28.5)L(t) = 0.25 - \sin\left(\frac{2\pi(t - 2)}{28.5}\right). This matches the general form with A=−1A = -1 (since the sine function is subtracted), B=2π28.5B = \frac{2\pi}{28.5}, C=2C = 2, and D=0.25D = 0.25.
  3. Calculate Period: Calculate the period of the function.\newlineUsing the relationship P=2π/BP = 2\pi/B, we can find the period of the given function by plugging in the value of BB from the function.\newlineP=2π/(2π/28.5)=28.5P = 2\pi / (2\pi/28.5) = 28.5

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Posted 10 months ago

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Posted 10 months ago

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Let g(x)=5−ex g(x)=5-e^{x} .\newlineBelow is Sean's attempt to write a formal justification for the fact that the equation g(x)=0 g(x)=0 has a solution where 1≤x≤4 1 \leq x \leq 4 .\newlineIs Sean's justification complete? If not, why?\newlineSean's justification:\newlineg g is defined for all real numbers, and exponential functions are continuous at all points in their domains.\newlineSo, according to the intermediate value theorem, g(x)=0 g(x)=0 must have a solution somewhere between x=1 x=1 and x=4 x=4 .\newlineChoose 11 answer:\newline(A) Yes, Sean's justification is complete.\newline(B) No, Sean didn't establish that 00 is between g(1) g(1) and g(4) g(4) .\newline(C) No, Sean didn't establish that g g is continuous.
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Posted 9 months ago

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Let f(x)=cos⁡(π⋅x)3 f(x)=\sqrt[3]{\cos (\pi \cdot x)} .\newlineBelow is Issac's attempt to write a formal justification for the fact that the equation f(x)=0.1 f(x)=0.1 has a solution where 1.5≤x≤2 1.5 \leq x \leq 2 .\newlineIs Issac's justification complete? If not, why?\newlineIssac's justification:\newlinef(1.5)=0 and f(2)=1, so f(1.5)<0.1<f(2). \begin{array}{l} f(1.5)=0 \text { and } \\ f(2)=1 \text {, so } \\ f(1.5)<0.1<f(2) . \end{array} \newlineSo, according to the intermediate value theorem, f(x)=0.1 f(x)=0.1 must have a solution when x x is between x=1.5 x=1.5 and x=2 x=2 .\newlineChoose 11 answer:\newline(A) Yes, Issac's justification is complete.\newline(B) No, Issac didn't establish that 00.11 is between f(1.5) f(1.5) and f(2) f(2) .\newline(C) No, Issac didn't establish that f f is continuous.
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Posted 9 months ago

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Let h(x)=5⋅tan⁡(x) h(x)=5 \cdot \tan (x) .\newlineBelow is Bridget's attempt to write a formal justification for the fact that the equation h(x)=2 h(x)=2 has a solution where 0≤x≤π4 0 \leq x \leq \frac{\pi}{4} .\newlineIs Bridget's justification complete? If not, why?\newlineBridget's justification:\newlineh h is defined over the entire interval [0,π4] \left[0, \frac{\pi}{4}\right] , and trigonometric functions are continuous at all points in their domains.\newlineSo, according to the intermediate value theorem, h(x)=2 h(x)=2 must have a solution at some point in that interval.\newlineChoose 11 answer:\newline(A) Yes, Bridget's justification is complete.\newline(B) No, Bridget didn't establish that 22 is between h(0) h(0) and h(π4) h\left(\frac{\pi}{4}\right) .\newline(C) No, Bridget didn't establish that h h is continuous.
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Posted 9 months ago

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Posted 9 months ago

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Let g(x)=cos⁡(πx2) g(x)=\cos \left(\pi x^{2}\right) .\newlineBelow is Amrita's attempt to write a formal justification for the fact that the equation g′(x)=0.4 g^{\prime}(x)=0.4 has a solution where −1<x<4 -1<x<4 .\newlineIs Amrita's justification complete?\newlineIf not, why?\newlineAmrita's justification:\newlinePolynomial and trigonometric functions are differentiable and continuous at all points in their domain, and [−1,4] [-1,4] is within g′ g^{\prime} s domain. Also,\newlineg(−1)=−1 and g(4)=1, so g(4)−g(−1)4−(−1)=0.4. \begin{array}{l} g(-1)=-1 \text { and } \\ g(4)=1 \text {, so } \\ \frac{g(4)-g(-1)}{4-(-1)}=0.4 . \end{array} \newlineSo, according to the mean value theorem, g′(x)=0.4 g^{\prime}(x)=0.4 must have a solution somewhere between x=−1 x=-1 and x=4 x=4 .\newlineChoose 11 answer:\newline(A) Yes, Amrita's justification is complete.\newline(B) No, Amrita didn't establish that the average rate of change of g g over [−1,4] [-1,4] is equal to 00.44 .\newline(C) No, Amrita didn't establish that g g is differentiable.
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Posted 9 months ago

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Let\newlineh(x)=x5−2x4+10x2−10x.  h(x)=x^{5}-2 x^{4}+10 x^{2}-10 x \text {. } \newlineBelow is Omar's attempt to write a formal justification for the fact that there exists a value c c where\newline−2<c<2 such that h′(c)=6.  -2<c<2 \text { such that } h^{\prime}(c)=6 \text {. } \newlineIs Omar's justification complete? If not, why?\newlineOmar's justification:\newlineh(−2)=−4 and h(2)=20, so h(2)−h(−2)2−(−2)=6. \begin{array}{l} h(-2)=-4 \text { and } \\ h(2)=20 \text {, so } \\ \frac{h(2)-h(-2)}{2-(-2)}=6 . \end{array} \newlineSo, according to the mean value theorem, there exists a value c c somewhere in the interval −2<c<2 -2<c<2 such that h′(c)=6 h^{\prime}(c)=6 .\newlineChoose 11 answer:\newline(A) Yes, Omar's justification is complete.\newline(B) No, Omar didn't establish that the average rate of change of h h over [−2,2] [-2,2] is equal to 66 .\newline(C) No, Omar didn't establish that h h is differentiable.
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Posted 9 months ago

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Let f(x)=x+x3 f(x)=\sqrt{x+\sqrt{x^{3}}} .\newlineBelow is Lena's attempt to write a formal justification for the fact that the equation f′(x)=23 f^{\prime}(x)=\frac{2}{3} has a solution where 0<x<9 0<x<9 .\newlineIs Lena's justification complete? If not, why?\newlineLena's justification:\newlinef(0)=0 f(0)=0 and f(9)=6 f(9)=6 , so the average rate of change of f f over the interval 0≤x≤9 0 \leq x \leq 9 is 23 \frac{2}{3} .\newlineSo, according to the mean value theorem, f′(x)=23 f^{\prime}(x)=\frac{2}{3} must have a solution somewhere in the interval 0<x<9 0<x<9 .\newlineChoose 11 answer:\newline(A) Yes, Lena's justification is complete.\newline(B) No, Lena didn't establish that the average rate of change of f f over[0,9] [0,9] is equal to 23 \frac{2}{3} .\newline(C) No, Lena didn't establish that f f is differentiable.
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Posted 9 months ago

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Let h(x)=3x−x2 h(x)=3^{x}-x^{2} .\newlineBelow is Uriah's attempt to write a formal justification for the fact that the equation h′(x)=8 h^{\prime}(x)=8 has a solution where 1<x<3 1<x<3 .\newlineIs Uriah's justification complete? If not, why?\newlineUriah's justification:\newlinePolynomial and exponential functions are differentiable and continuous at all points in their domain, and [1,3] [1,3] is within h h 's domain. Also, h(1)=2 h(1)=2 and h(3)=18 h(3)=18 , so\newlineh(3)−h(1)3−1=8.  \frac{h(3)-h(1)}{3-1}=8 \text {. } \newlineSo, according to the mean value theorem, h′(x)=8 h^{\prime}(x)=8 must have a solution somewhere between x=1 x=1 and x=3 x=3 .\newlineChoose 11 answer:\newline(A) Yes, Uriah's justification is complete.\newline(B) No, Uriah didn't establish that the average rate of change of h h over [1,3] [1,3] is equal to 88 .\newline(C) No, Uriah didn't establish that h h is differentiable.
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