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Rainbow [258]
3 years ago
6

How many perfect squares are there between 2 and 140​

Mathematics
1 answer:
zlopas [31]3 years ago
8 0

Answer:

11

Step-by-step explanation:

4, 9, 16, 25, 36, 49, 64, 81, 100, 121

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A vehicle travels a distance of 5t^4-10t^2+6 miles in t+2 minutes. What is the vehicles speed in miles/minutes?
8090 [49]

Answer: \left(5t^4-10t^2+6\right)\cdot \:l\cdot \:e^2\cdot \:i^2\cdot \:n\cdot \:t\cdot \left(t+2\right)\cdot \:s^2\cdot \:u

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4 0
3 years ago
Choose any method to find the gcf of 40 16 and 24
Juli2301 [7.4K]

The GCF of 40 16 and 24 is 8

<h3>How to determine the GCF of 40 16 and 24?</h3>

The numbers are given as:

40 16 and 24

Start by listing out the factors of the numbers:

This is done as follows:

  • The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40
  • The factors of 16 are: 1, 2, 4, 8, 16
  • The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24

Then the greatest common factor in the above list is 8.

Hence, the GCF of 40 16 and 24 is 8

Read more about GCF at:

brainly.com/question/219464

#SPJ1

8 0
2 years ago
Help me PLEASEEEEEEEEEEEEE
tatuchka [14]

Answer:

Step-by-step explanation:

A, D and F

4 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
Bigideasmath.
ratelena [41]
The total number of calories for 18 cups of milk is 1620 calories.
If two cups of milk is 180 calories then divide 180/2 to find the unit rate.
Then multiply 90 by the number of cups (18) and then you get the total number of calories for that many drinks.
6 0
3 years ago
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