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gulaghasi [49]
3 years ago
15

Sue walked from her home to her school at a constant rate of 2 miles per hour, picked up her bicycle and rode back at a constant

rate of 10 miles per hour. If the round trip took 1.5 hours, how far is the school from Sue's home?
Mathematics
2 answers:
Nady [450]3 years ago
8 0

Answer:

15 miles

Step-by-step explanation:

the 2 mph is a red herring. If there is a rate of 10 mph, and the trip took 1.5 hours, the 1 hour=10 miles, and the .5=5 miles, add em up and its 15 miles

dem82 [27]3 years ago
8 0

Answer: the school is 2.5 miles from Sue's home.

Step-by-step explanation:

Let x represent the distance of the school from Sue's home.

Time = distance/speed

Sue walked from her home to her school at a constant rate of 2 miles per hour. It means that the time it took her to walk to the school is

x/2

She picked up her bicycle and rode back at a constant rate of 10 miles per hour. The distance covered is the same and the time spent in riding back home is

x/10

If the round trip took 1.5 hours, it means that

x/2 + x/10 = 1.5

Multiplying through by 10, it becomes

5x + x = 15

6x = 15

x = 15/6

x = 2.5 miles

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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}

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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.

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PLESE HELP ill GIVE BRANILY<br><br> Solve for x. 3.5(7-x)+32=106-1.5(4x+18)
murzikaleks [220]

Answer:

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Step-by-step explanation:

3.5(7−x)+32=106−1.5(4x+18)

Step 1: Simplify both sides of the equation.

3.5(7−x)+32=106−1.5(4x+18)

(3.5)(7)+(3.5)(−x)+32=106+(−1.5)(4x)+(−1.5)(18)(Distribute)

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−3.5x+56.5=−6x+79

Step 2: Add 6x to both sides.

−3.5x+56.5+6x=−6x+79+6x

2.5x+56.5=79

Step 3: Subtract 56.5 from both sides.

2.5x+56.5−56.5=79−56.5

2.5x=22.5

Step 4: Divide both sides by 2.5.

2.5x

2.5

=

22.5

2.5

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