Here is a list of the odd number paired

1+3, 1+5, !+7, and !+9 (there are 4 unique sums - 4, 6,8 and 10)
3+5, 3+7, 3+9 (notice I did not pair 3 with 1 and the the only new sum is 12)
5+7, 5+9 (the only new sum is 14)
7+9 (16 is a new sum)

The sums (no repeats) are 4,6,8,10,12,14 and 16 for a total of seven numbers.

5 0

3 years ago

Guys help please I am struggling

alina1380 [7]

9514 1404 393

Answer:

14.1 years

Step-by-step explanation:

Use the compound interest formula and solve for t. Logarithms are involved.

A = P(1 +r/n)^(nt)

amount when P is invested for t years at annual rate r compounded n times per year.

Using the given values, we have ...

13060 = 8800(1 +0.028/365)^(365t)

13060/8800 = (1 +0.028/365)^(365t) . . . . divide by P=8800

Now we take logarithms to make this a linear equation.

log(13060/8800) = (365t)log(1 +0.028/365)

Dividing by the coefficient of t gives us ...

t = log(13060/8800)/(365·log(1 +0.028/365)) ≈ 0.171461/0.0121598

t ≈ 14.1

It would take about 14.1 years for the value to reach $13,060.

8 0

3 years ago

Help pleaseeeeeeeeeeeeeeeeeeeeeee

bixtya [17]

Answer: $\bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}$

<u>Step-by-step explanation:</u>

(1) (12, 18, 27, ...)

The common ratio is:

$r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\ r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}$

$(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\ r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}$

3 0

3 years ago