GEOMETRY 100 POINTS CHALLENGE

find x​

Answer:

[tex]\dfrac{5e^2}{2}[/tex]

Step-by-step explanation:

Differentiation is an algebraic process that finds the slope of a curve. At a point, the slope of a curve is the same as the slope of the tangent line to the curve at that point. Therefore, to find the slope of the line tangent to the given function, differentiate the given function.

Given function:

[tex]y=x^2\ln(2x)[/tex]

Differentiate the given function using the product rule.

[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]

[tex]\textsf{Let\;$u=x^2}[/tex][tex]\textsf{Let\;$u=x^2$}\implies \dfrac{\text{d}u}{\text{d}x}=2x[/tex]

[tex]\textsf{Let\;$v=\ln(2x)$}\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{2}{2x}=\dfrac{1}{x}[/tex]

Input the values into the product rule to differentiate the function:

[tex]\begin{aligned}\dfrac{\text{d}y}{\text{d}x}&=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}\\\\&=x^2 \cdot \dfrac{1}{x}+\ln(2x) \cdot 2x\\\\&=x+2x\ln(2x)\end{aligned}[/tex]

To find the slope of the tangent line at x = e²/2, substitute x = e²/2 into the differentiated function:

[tex]\begin{aligned}x=\dfrac{e^2}{2}\implies \dfrac{\text{d}y}{\text{d}x}&=\dfrac{e^2}{2}+2\left(\dfrac{e^2}{2}\right)\ln\left(2 \cdot \dfrac{e^2}{2}\right)\\\\&=\dfrac{e^2}{2}+e^2\ln\left(e^2\right)\\\\&=\dfrac{e^2}{2}+2e^2\\\\&=\dfrac{5e^2}{2}\end{aligned}[/tex]

Therefore, the slope of the line tangent to the graph of y = x²ln(2x) at the point where x = e²/2 is:

[tex]\boxed{\dfrac{5e^2}{2}}[/tex]

The annualized return for the entire period is the closest to 10.5%.

To calculate the annualized return for the entire period, we need to consider the returns for each year and the return in the first quarter of Year 4. Since the returns are given for each period, we can use the geometric mean to calculate the annualized return.

The formula for calculating the geometric mean return is:

Geometric Mean Return = [(1 + R1) * (1 + R2) * (1 + R3) * (1 + R4)]^(1/n) - 1

Where R1, R2, R3, and R4 are the returns for each respective period, and n is the number of periods.

Given the returns:

Year 1 return: 5% or 0.05

Year 2 return: 12% or 0.12

Year 3 return: -6% or -0.06

First quarter of Year 4 return: 2% or 0.02

Using the formula, we can calculate the annualized return:

Annualized Return = [(1 + 0.05) * (1 + 0.12) * (1 - 0.06) * (1 + 0.02)]^(1/3) - 1

Annualized Return = (1.05 * 1.12 * 0.94 * 1.02)^(1/3) - 1

Annualized Return = 1.121485^(1/3) - 1

Annualized Return ≈ 0.105 or 10.5%

Therefore, the annualized return for the entire period is approximately 10.5%.

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The average rate of change of the school's enrollment during this time period is -49 students per year. This means that on average, the enrollment at the college has been decreasing by 49 students per year.

To determine the average rate of change of the school's enrollment during the given time period, we can use the formula:

Average rate of change = (Change in enrollment) / (Change in time)

The change in enrollment is calculated by subtracting the initial enrollment from the final enrollment, while the change in time is calculated by subtracting the initial year from the final year.

Given that in 2004 there were 975 students enrolled and in 2009 there were 730 students enrolled, we can calculate the change in enrollment:

Change in enrollment = 730 - 975 = -245 students

The change in time can be calculated as:

Change in time = 2009 - 2004 = 5 years

Now we can calculate the average rate of change:

Average rate of change = (-245 students) / (5 years) = -49 students per year

Therefore, the average rate of change of the school's enrollment during this time period is -49 students per year. This means that on average, the enrollment at the college has been decreasing by 49 students per year.

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