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irinina [24]
3 years ago
7

The area of a book cover is 63.8 in². The width of the book is the difference of the length and 3 inches. Let x represent the bo

ok's length. Which quadratic equation represents the area of the book cover?
x(x - 3) = 63.8
x(3 - x) = 63.8
2x - 3 = 63.8
x + x - 3 = 63.8
Mathematics
1 answer:
Mazyrski [523]3 years ago
3 0
In this item, we are already given that the length of the rectangular book can be represented by the variable x. In this way, the width, which was described to be the difference of the length and 3 inches, can be expressed as x - 3. 

The area of the triangle is rectangle is calculated by multiplying the length, x, and the width, which was derived to be x - 3.

                 Area = Length x width
                 Area = (x)( x - 3)

Putting in the value of the area, we have the final answer as,
               x(x - 3) = 63.8 in²

Hence, the final answer is the first answer. 
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Be sure to answer all parts. List the evaluation points corresponding to the midpoint of each subinterval to three decimal place
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Answer:

The Riemann Sum for \int\limits^5_4 {x^2+4} \, dx with n = 4 using midpoints is about 24.328125.

Step-by-step explanation:

We want to find the Riemann Sum for \int\limits^5_4 {x^2+4} \, dx with n = 4 using midpoints.

The Midpoint Sum uses the midpoints of a sub-interval:

\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)

where \Delta{x}=\frac{b-a}{n}

We know that a = 4, b = 5, n = 4.

Therefore, \Delta{x}=\frac{5-4}{4}=\frac{1}{4}

Divide the interval [4, 5] into n = 4 sub-intervals of length \Delta{x}=\frac{1}{4}

\left[4, \frac{17}{4}\right], \left[\frac{17}{4}, \frac{9}{2}\right], \left[\frac{9}{2}, \frac{19}{4}\right], \left[\frac{19}{4}, 5\right]

Now, we just evaluate the function at the midpoints:

f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(4\right)+\left(\frac{17}{4}\right)}{2}\right)=f\left(\frac{33}{8}\right)=\frac{1345}{64}=21.015625

f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{17}{4}\right)+\left(\frac{9}{2}\right)}{2}\right)=f\left(\frac{35}{8}\right)=\frac{1481}{64}=23.140625

f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(\frac{9}{2}\right)+\left(\frac{19}{4}\right)}{2}\right)=f\left(\frac{37}{8}\right)=\frac{1625}{64}=25.390625

f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{19}{4}\right)+\left(5\right)}{2}\right)=f\left(\frac{39}{8}\right)=\frac{1777}{64}=27.765625

Finally, use the Midpoint Sum formula

\frac{1}{4}(21.015625+23.140625+25.390625+27.765625)=24.328125

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C is the best solution.

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