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2 years ago

Recursive formula for a geometric sequence with a starting value of 8 and a common ratio of 5

1 answer:

5 0

$a_2=5a_1$
$a_3=5a_2$
$a_4=5a_3$
...
$a_n=5a_{n-1}$

So the recursive formula is

$\begin{cases}a_1=8\\a_n=5a_{n-1}&\text{for }n>1\end{cases}$

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Will mark brainlist if its right

pickupchik [31]

Answer: S<\-2

Hope i was some help.

Step-by-step explanation: sorry I can’t put the right sign my keybored won’t let me but you know what it stands for.

3s+6s</-5(s+2)

start off with -5(s+2) you multiply -5 with s and it is -5s then you do the same for +2 take -5 and multiply by +2 pos times a neg is always a neg so it’s -10.

3s+6<\-5s-10

So you have to get rid of the -5s so your going to do the inverse which is add 5 to -5 and it will get rid of -5s but then you have to add 5 to 3s and that will be 8s.

8s+6</-10

Now your going to get rid of the +6 and to do that you do the inverse you subtract 6 and it will get rid of it now you have to do the same for -10 so you will subtract 6 to -10 and that will equal -16.

Now you have to divide!

8s/8 </ -16/8

And it should equal s <\-2

3 0

2 years ago

Help me with this question

Daniel [21]

Answer: A) 23,00

Step-by-step explanation:

A=23,00

B=8,600

C=12,00

D=3,600

3 0

1 year ago

Read 2 more answers

Find the roots of:

Elena-2011 [213]

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{1) \ 2x^3-7x^2+8x-3=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -3 are 1, -1, 3, +3. So:

| 2 -7 8 -3

1 | 2 -5 3

| 2 -5 3 0

1 | 2 -3

2 -3 0

So the factorization is (x-1)² (2x-3)=0. So:

$\bf{ x_1=x_2=1 \qquad x_2=\dfrac{3}{2} }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{2) \ x^3-x^2-4=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of -4 are 1, -1, 2, -2, 4, -4. So:

| 1 -1 0 -4

2 | 2 2

1 2 2 0

So the factorization is (x-2)(x²+x+2)=0 . When calculating the discriminant of the trinomial, it is concluded that it has no roots since the result is negative. So you only have one solution.

$\bf{ 1^2-4(2)(2)=1-16=-15 < 0 \quad \Longrightarrow \quad x=2 }$

$\large\displaystyle\text{$\begin{gathered}\sf \pmb{3) \ 6x^3+7x^2-9x+2=0 } \end{gathered}$}$

Synthetic division is used since the equation is of the third degree. The divisors of 2 are 1, -1, 2, -2. So:

| 6 7 9 2

-2 | -12 10 -2

6 -5 1 0

So the factorization is (x+2)(6x²-5x+1)=0 . The quadratic equation is solved by the general formula:

$\bf{ x_{2, 3}&=\dfrac{5\pm \sqrt{(5)^2-4(6)(1)}}{2(6)}=\dfrac{5\pm \sqrt{25-24}}{12}=\dfrac{5\pm 1}{12} }}$

$\large\displaystyle\text{$\begin{gathered}\sf \begin{matrix} x_1=-2&\ \ \ \ \ \ x_{2}=\dfrac{6}{12} \qquad &\ \ \ x_3=\dfrac{4}{12}\\ &\ \ \ x_2=\dfrac{1}{2} \qquad &x_3=\dfrac{1}{3} \end{matrix} \end{gathered}$}$

6 0

1 year ago

The cost for n teachers to attend the technology conference is 3n+5 dollars. What is the cost for 7 teachers to attend

Veronika [31]

If it costs 3n+5, you have to plug 7 in for n and solve. 3(7)+5=26 so $26

3 0

2 years ago

I need help translating expressions and equations 5n-4=12

azamat

Answer: 16/5

5n=12+4
5n=16
n= 16/5

6 0

2 years ago

Read 2 more answers