To compare their speed (rate), we need to have the same time or same distance to use as a reference.

Matt ran 50 meters per 10 seconds. Molly ran 30 meters per 5 seconds. If she ran for 5 more seconds, she would have gone 30 more meters, for 30 meters per 10 seconds. Thus, Molly went faster.

Alternately, you could calculate their speeds in meters per (one) second and compare. 50 meters per 10 seconds is 5 meters per second, and 30 meters per 5 seconds is 6 meters per second.

8 0

3 years ago

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What is the sum of the first 51 consecutive odd positive integers?

Angelina_Jolie [31]

We call:

$a_{n}$ as the set of the first 51 consecutive odd positive integers, so:

 $a_{n} = \{1, 3, 5, 7, 9...\}$

Where:
$a_{1} = 1$
$a_{2} = 3$
$a_{3} = 5$
$a_{4} = 7$
$a_{5} = 9$
and so on.

In mathematics, a sequence of numbers, such that the difference between two consecutive terms is constant, is called Arithmetic Progression, so:

3-1 = 2
5-3 = 2
7-5 = 2
9-7 = 2 and so on.

Then, the common difference is 2, thus:

 $a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\}$

Then:

 $a_{n} = a_{1} + (n-1)d$

So, we need to find the sum of the members of the finite series, which is called arithmetic series:

There is a formula for arithmetic series, namely:

 $S_{k} = ( \frac{a_{1} + a_{k}}{2} ).k$

Therefore, we need to find:
 $a_{k} = a_{51}$

Given that $a_{1} = 1$ , then:

$a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1$

Thus:
$a_{k} = a_{51} = 2(51)-1 = 101$

Lastly:

$S_{51} = ( \frac{1 + 101}{2} ).51 = 2601$

4 0

3 years ago