GEOMETRY 100 POINTS

FIND THE VALUE OF X​

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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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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Answer:

x = 23

Step-by-step explanation:

The diagonals of a rectangle are equal in size:

2x+10 = 56 (subtract 10 from both sides)

2x = 46 (divide by 2 for both sides)

x = 23

So x = 23

Answer:

x = 6

Step-by-step explanation:

The diagonals of a Rhombus bisect the angles

⇒ 10x - 23 = 3x + 19

⇒ 10x - 3x = 19 + 23

⇒ 7x = 42

⇒ x = 6

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 square root of 76.8 is 8.76.

Square root of 76.8 is 8.746.

Given,

[tex]\sqrt{76.8}[/tex]

Now,

Simplifying the square root :

If square root of x is to be calculated then,

[tex]\sqrt{x}[/tex] = [tex]x^{1/2}[/tex]

Similarly,

[tex]\sqrt{76.8}[/tex] = [tex]76.8^{1/2}[/tex]

= 8.746

Thus the square root of 76.8 is 8.746.

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Answer:

[tex](-\infty,2)[/tex]

Step-by-step explanation:

Since [tex]-3x+6\nless 0[/tex], then [tex]x\ngtr 2[/tex], therefore, the domain of the function is [tex](-\infty,2)[/tex].

answer: 190%

step-by-step explanation:

hihi your problem is to find the percent increase. we can use the following formula!

percent Increase = ((new value - old value) / old value) * 100

given that bruce's initial typing speed was 10 words per minute (old value) and his final typing speed after the course was 29 words per minute (new value), we can substitute these values into the formula:

percent increase = ((29 - 10) / 10) * 100

simplifying the numerator:

percent increase = (19 / 10) * 100

dividing 19 by 10:

percent increase = 1.9 * 100

calculating the product:

percent Increase = 190

therefore, the percent increase in Bruce's typing speed after taking the course is 190%. this means that his typing speed improved by 190% compared to his initial speed.

hopefully this helped !!