#3 what is 33/222 simple form

Mathematics

1 answer:

vfiekz [6]3 years ago

5 0

Answer: 11/74

Step-by-step explanation:

We have to think about what number can be divided from both 33 and 222. 33’s factors are 11 and 3. 222’s are 3, 6, 74, 37, and probably more. The only one the two numbered have in common is 3. So, divide north the numbers toe and denominator by 3 to get 11/74.

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A group of adults and students went on a class trip to Washington, DC. The number of male students was 1 more than 7 times the n

sp2606 [1]

Let the number of adults be x, then number of male students is 1 + 7x and the number of female students is (1 + 7x)/2.
x + 1 + 7x + (1 + 7x)/2 = 82
2x + 2 + 14x + 1 + 7x = 164
23x = 164 - 2 - 1 = 161
x = 161/23 = 7

Therefore, number of male students is 1 + 7(7) = 1 + 49 = 50
and the number of female students is 50/2 = 25.

5 0

3 years ago

Help? I can't seem to understand arithmegic sequence.

katrin2010 [14]

A sequence $\{a_n\}$ is arithmetic if the difference between consecutive terms is some fixed number, regardless of which pair of consecutive terms you pick out of the sequence.

For example, the following sequences are arithmetic:

1, 2, 3, 4, 5, 6, ... (difference = 1)

-25, -20, -15, -10, -5, ... (difference = 5)

2. Carla's sequence is not arithmetic, because the differences between consecutive terms are all different:

13 - 11 = 2

17 - 13 = 4

25 - 17 = 8

She can adjust the sequence by changing the last two numbers to 15 and 17, since this makes the difference fixed:

13 - 11 = 2

15 - 13 = 2

17 - 15 = 2

and so on.

3. The sequence

45, 48, 51, 54, ...

is arithmetic with difference 3 between terms. Recursively, we can write the $n$ th term, $a_n$ , in terms of the previous, $(n-1)$ th term, $a_{n-1}$ :

$a_n=a_{n-1}+3$

By this definition, we can just as easily write the $(n-1)$ th term in terms of the $(n-2)$ th term:

$a_{n-1}=a_{n-2}+3$

Then, substituting this into the previous equation, we have

$a_n=(a_{n-2}+3)+3=a_{n-2}+2\cdot3$

We can continue this process to write $a_n$ in terms of $a_1$ :

$a_{n-2}=a_{n-3}+3\implies a_n=a_{n-3}+3\cdot3$

$a_{n-3}=a_{n-4}+3\implies a_n=a_{n-4}+4\cdot3$

and so on. (You might notice that the subscript of the term on the right side, and the number of 3s being added, together sum to $n$ .) The pattern continues down to

$a_n=a_1+(n-1)\cdot3$