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solmaris [256]
3 years ago
5

Answer #8 for 27 points and BRAINLIEST ANSWER. SOLVE QUICKLY PLEASE!!!

Mathematics
2 answers:
Semmy [17]3 years ago
7 0
A and B will use the formulas: 4/3 * pi * r^3 

A = 8.181 in^3

B = 9.471 in^3

C write an inequality

8.181 < x < 9.471 in^3
Softa [21]3 years ago
4 0
The agree with tanner23456
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A video game was originally priced at $50, but Tess waited to buy it until the video game was
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$21

Step-by-step explanation:

$50×.60=$30

$50-$30=$20

$20×.05=$1

$20+$1=$21

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3 years ago
2. State the domain for the following ordered pairs: { (4, -2), (6, -7), (4, 3), (-8, 1), (-3, 5) } *
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Answer:

(4,6,4,-8,-3)

Step-by-step explanation:

Domain means of function is the set of all possible inputs for function

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28. Katrina buys a 64-ft roll of fencing to
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Janet gets paid $12 per hour babysitting for h hours.
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Step-by-step explanation:

3 0
3 years ago
Be sure to answer all parts. List the evaluation points corresponding to the midpoint of each subinterval to three decimal place
gayaneshka [121]

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

This is the sketch of the function and the approximating rectangles.

5 0
4 years ago
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