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Lostsunrise [7]

3 years ago

15

The sum of two numbers is fourteen. One number is two less than three times the other. Find the numbers.

1 answer:

Margarita [4]3 years ago

7 0

So, for the first variable, you would use x, and for the second, y. Then, you would write out 2-(3y)=14. once this is done, youd simplify to 2-3y=14. Subtract 2 from both sides to get 3y=12. Then, divide both sides by 3. this will leave you with y=4. From this you can see the other number is ten. Hope this helped! and i apologize if it is incorrect!

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Answer:

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The $n^{th}$ term of geometric sequence is given by:

$a_n = ar^{n-1}$

$a_4 = 16 = ar^3\\a_6 = 4 = ar^5$

Dividing the two equations, we get,

$\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}$

the first term can be calculated as:

$16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128$

Thus, the required geometric sequence is

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

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