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2 years ago

: the mass of an average classroom meter stick is roughly 0. 2 kg.

Mathematics

1 answer:

tatiyna2 years ago

8 0

Using proportions, it is found that it takes $3.5 \times 10^{23}$ meter sticks to equal the mass of the moon.

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

In this problem, one meter stick has a mass of 0.2 kg. How many meter sticks are needed for a mass of $7 \times 10^{22} \text{kg}$ ? The rule of three is given by:

One meter stick - 0.2 kg

x meter sticks - $7 \times 10^{22} \text{kg}$

Applying cross multiplication:

$0.2x = 7 \times 10^{22}$

$x = \frac{7 \times 10^{22}}{0.2}$

$x = 35 \times 10^{22}$

$x = 3.5 \times 10^{23}$

It takes $3.5 \times 10^{23}$ meter sticks to equal the mass of the moon.

More can be learned about proportions at brainly.com/question/24372153

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The solution would be like this for this specific problem:

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3 years ago

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bija089 [108]

Answers:

$t_{10} = -22 \ \text{ and } S_{10} = -85$

========================================================

Explanation:

$t_1 = \text{first term} = 5\\t_2 = \text{first term}-3 = t_1 - 3 = 5-3 = 2$

Note we subtract 3 off the previous term (t1) to get the next term (t2). Each new successive term is found this way

$t_3 = t_2 - 3 = 2-3 = -1\\t_4 = t_3 - 3 = -1-3 = -4$

and so on. This process may take a while to reach $t_{10}$

There's a shortcut. The nth term of any arithmetic sequence is

$t_n = t_1+d(n-1)$

We plug in $t_1 = 5 \text{ and } d = -3$ and simplify

$t_n = t_1+d(n-1)\\t_n = 5+(-3)(n-1)\\t_n = 5-3n+3\\t_n = -3n+8$

Then we can plug in various positive whole numbers for n to find the corresponding $t_n$ value. For example, plug in n = 2

$t_n = -3n+8\\t_2 = -3*2+8\\t_2 = -6+8\\t_2 = 2$

which matches with the second term we found earlier. And,

$tn = -3n+8\\t_{10} = -3*10+8\\t_{10} = -30+8\\t_{10} = \boldsymbol{-22} \ \textbf{ is the tenth term}$

---------------------

The notation $S_{10}$ refers to the sum of the first ten terms $t_1, t_2, \ldots, t_9, t_{10}$

We could use either the long way or the shortcut above to find all $t_1$ through $t_{10}$ . Then add those values up. Or we can take this shortcut below.

$Sn = \text{sum of the first n terms of an arithmetic sequence}\\S_n = (n/2)*(t_1+t_n)\\S_{10} = (10/2)*(t_1+t_{10})\\S_{10} = (10/2)*(5-22)\\S_{10} = 5*(-17)\\\boldsymbol{S_{10} = -85}$

The sum of the first ten terms is -85

-----------------------

As a check for $S_{10}$ , here are the first ten terms:

t1 = 5
t2 = 2
t3 = -1
t4 = -4
t5 = -7
t6 = -10
t7 = -13
t8 = -16
t9 = -19
t10 = -22

Then adding said terms gets us...

5 + 2 + (-1) + (-4) + (-7) + (-10) + (-13) + (-16) + (-19) + (-22) = -85

This confirms that $S_{10} = -85$ is correct.

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