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givi [52]
3 years ago
9

Help please it's overdue:(

Mathematics
1 answer:
bearhunter [10]3 years ago
8 0
What grade is this for by the way?
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Determine whether the sequence is geometric. If so, find the common ratio. 3.0, 6.0, 12.0, 24.0
Luba_88 [7]

Answer:

The sequence <u>is</u> geometric and the common ratio is 2

Step-by-step explanation:

A sequence is geometric if each term  (after the first one) is equal to the term before it times some constant number, called the 'common ratio'.

Examples:  1, 2, 4, 8, 16, ... has a common ratio of 2 because 2*2 = 4, and 4*2 = 8, and so on.

                 Also, -->  4, 2, 1, 1/2, 1/4/ 1/8, ... has a common ratio of 1/2 because

4 * 1/2 =  the next term 2, and 2 * 1/2 = the next trem 1/4, and son on.

In this case, 3 * 2 = 6

                    6 * 2 = 12

                    12 * 2 = 24

The sequence <u>is</u> geometric and the common ratio is 2

6 0
3 years ago
Convert 65 mm to meters
Ymorist [56]
0.065 meters is equal to 65 millimeters
7 0
4 years ago
Answer plzzzz I need it ok
Mariana [72]

Answer:

There are 26 different possible outcomes. the probability of the computer choosing the first letter of your name is 1 out of 26.

4 0
3 years ago
What is proppotion of 10/15 =12/x?
Marianna [84]
12 divided by 10 is 1.2. 15 multiplied by 1.2 is 18 so x equals 18
7 0
3 years ago
Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th
xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is \bold{W=-3e^{7x}}

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|

Thus replacing the functions of the exercise we get:

W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|

Working with the determinant we get

W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0
3 years ago
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