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grigory [225]
3 years ago
12

math Juan and Brooke measured a paper clip Juan said the paper clip was 5 inches and Brooke said it was 2 inches how could both

students be correct.
Mathematics
1 answer:
aleksklad [387]3 years ago
8 0
They could both be correct, since one could be measuring the width and the other the height. Typically a paper clip has a bigger length than width, so Juan most likely measured the length and Brooke probably measured the width.
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The lowest term of 836/54 is 418/27!
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The rule for calculating the mean is to add up all the scores in a sample and divide by the _________
Tems11 [23]
Amount of numbers in the sample.
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oliver is training for a marathon. In practice, he runs 15 kilometers in 72 minutes. What is the speed in kilometers per hour?
34kurt
72/15=4.8
4.8 is the answer 
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4 years ago
Angle A and angle B are complementary angles. If m∡a = 4x and m∡b = 3x +13, what is the measure of the smaller angle?
bagirrra123 [75]

Answer:

Step-by-step explanation:

Complementary angles mean two two angles sum with equal 90 degrees. Therefore you would need to create an equation to solve for the value of x.

4x+3x+13=90

-13 -13

7x=77

/7 /7

X=11

Now plug in the value of x.

A=4(11) B=3(11)+13

A=44. B=33+13

B=46

Angle a is the smaller angle and measures at 44°

5 0
2 years ago
Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
Aleks [24]

Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

5 0
3 years ago
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