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shtirl [24]
3 years ago
7

PLEASE HELP ME ILL GIVE YOU BRINIEST

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
4 0

Answer:

it is the second one

Step-by-step explanation:

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an artist has collected 15/16 pounds of seashells. he wants to give away 8/21 of the total weight of the seashells. how many pou
Semmy [17]

\frac{15}{16} \times  \frac{8}{21} =  \frac{120}{336}  \\ =  \frac{5}{14} \: pounds
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CAN SOMEONE PLEASE HELP ME ASAP ILL MARK BRAINLIST !!!!
liubo4ka [24]

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yes

Step-by-step explanation:

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Write down the fifth term of the sequence that has its nth term:
Mice21 [21]

Answer:

a) 34

b) 30

Step-by-step explanation:

so the nth term is just the position on the term, so if u want to find the 5th term you just have to substitute 5 in for n and solve!

a) 6n + 4

= 6(5) + 4

= 30 + 4

= 34          --> the 5th term for a sequence defined by 6n+4 is 34

b) n^2 + 5

= (5)^2 + 5

= 25 + 5

= 30           --> the 5th term for a sequence defined by n^2 + 5 is 30

hope this helps!

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.

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4 years ago
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alexandr402 [8]
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Hope this helps:)<span />
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3 years ago
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