What is the quotient when 2x3 + x + 3 is divided by x + 1?

If the common ratio of a geometric sequence is less

Vilka [71]

Answer:

Step-by-step explanation:

it's going to alternate back and froth between positive and negative, maybe it will make one of those "telescoping" sequences

6 0

2 years ago

If Amelia would like to double her money in twelve years, how much interest does she need to earn? 6 percent 12 percent 10 perce

sveticcg [70]

We can assign the value "2x" as the amount of money that Amelia which to achieve in twelve years.
So, his initial money is X.
We can try each percent to test if it could double the amount in 12 years.
x+0.06*12years*x=1.72X
x+0.12*12*x=1.44x
x+0.1*12*x=2.2x
x+0.166*12*x=2.992x

Therefore, the answer is 10 percent.

4 0

3 years ago

When you add two odd numbers together, the answer is:

Mashcka [7]

Answer:

1. Even - B

2. Even - B

3. Odd - A

Step-by-step explanation:

Example 1: 7+7 = 14

7 is odd, 14 is even.

Example 2: 4 + 4 = 8

4 is even, and 8 is even.

Example 3: 4 + 5 = 9

4 is even, 5 is odd, 9 is odd.

3 0

3 years ago

Read 2 more answers

Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is