What is the answer to 9y-6y+y=

Mathematics

2 answers:

Vlada [557]2 years ago

8 0

The answer is: 4y
Because, 9y-6y = 3y + y = 4y

babunello [35]2 years ago

4 0

9y - 6y = 3y
3y + y = 4y

answer: 4y

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Step-by-step explanation:

Well since the questions are related to geometric series I'm assuming the reasoning for 0.99999.... = 1, is going to be a geometric series in this instance. All though there are other ways to prove this.

a. So if you think about it, 0.999 can be represented as: $\frac{9}{10} + \frac{9}{100} + \frac{9}{1000}....$ and so on. This is a geometric series. The geometric sequence explicit equation can be determined by realizing each term is the previous term multiplied by 1/10, since the numerator is staying the same, but the denominator is being multiplied by 1/10. So the explicit equation for the sequence is: $a_n=\frac{9}{10}*(\frac{1}{10})^{n-1}$ . The sum of an infinite series can be defined as: $S_\infty=\frac{a_1}{1-r}$ when |r| <1 , which in this case it is. So plugging the values into the equation you get: $S_\infty=\frac{\frac{9}{10}}{1-0.10}=\frac{\frac{9}{10}}{\frac{9}{10}} = 1$ . So the sum of this infinite series that represents $0.\overline{9}$ is equal to 1.

b. Another repeating decimal can easily be determined using the repeating decimal 0.9. If you multiply both sides by 10, you'll get: $9.\overline9 = 10$ .

a. So this is an arithmetic sequence and it appears to be increasing by 5, with the first term being 3. So an explicit formula can be written using these two values: $a_n=3+5(n-1)$ which can also be simplified by distributing the 5, to get: $a_n=3+5n-5=5n-2$ . A recursive rule can also be easily defined using these two values, it's simply: $a_n=a_{n-1}+5\\a_1=3$

b. The 1000th term can be evaluated using the explicit formula by plugging in 1000 in as n: $a_{1000} =5(1000)-2=5000-2=4998$

c. The explicit formula, since it's much easier to evaluate values, by simply plugging in 1000 as n

d. The issue with a recursive rule, is you need to know all previous values, to determine $a_n$ , because it's doing some operation on the previous value, except to know that previous value, you need the previous previous value, and so on. This is an issue with much larger numbers, which is why the explicit formula is generally more better.