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2 years ago

8

Find the first, fourth, and eighth terms of the sequence.

1 answer:

Oksana_A [137]2 years ago

6 0

$a(n)=-3\cdot2^{n}-1\\\\a_1=-3\cdot2^1-1=-3\cdot2-1=-6-1=-7\\a_4=-3\cdot2^4-1=-3\cdot16-1=-48-1=-49\\a_8=-3\cdot2^8-1=-3\cdot256-1=-768-1=-769\\\\if\ a(n)=-3\cdot2^{n-1}\ then\\\\a_1=-3\cdot2^{1-1}=-3\cdot2^0=-3\cdot1=-3\\a_4=-3\cdot2^{4-1}=-3\cdot2^3=-3\cdot8=-24\\a_8=-3\cdot2^{8-1}=-3\cdot2^7=-3\cdot128=-384$

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A researcher gathers a sample of n=89 people to try a new

muminat

Answer:

24

Step-by-step explanation:

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5 0

2 years ago

How many, and what type of, solutions does 4x² + 12x + 9 = 0 have? 2 irrational solutions 1 rational solution 2 nonreal solution

SVEN [57.7K]

The discriminant is

$\sqrt{ {12}^{2} - 4(4)(9)}$

which is 0.

Since the discriminant is 0, there is only one real solution.

4 0

3 years ago

Solve the following system of equations. What is one x-value of a solution?

gulaghasi [49]

Answer:

There is some mistake in the question, because the solutions are x = -1.445 and x = -34.555

Step-by-step explanation:

Given the functions:

f(x) = x² + 4x + 10

g(x) = -32x - 40

we want to find the points at which f(x) = g(x).

x² + 4x + 10 = -32x - 40

x² + 4x + 10 + 32x + 40 = 0

x² + 36x + 50 = 0

Using quadratic formula:

$x = \frac{-b \pm \sqrt{b^2-4(a)(c)}}{2(a)}$

$x = \frac{-36 \pm \sqrt{36^2-4(1)(50)}}{2(1)}$

$x = \frac{-36 \pm 33.11}{2}$

$x_1 = \frac{-36 + 33.11}{2}$

$x_1 = -1.445$

$x_2 = \frac{-36 - 33.11}{2}$

$x_2 = -34.555$

5 0

2 years ago

30 POINTS NO CAP PLEASE HELP ME NEED RIGHT ANSWER

Cloud [144]

I think it’s 14/-28 which simplifies to 7/-14 which then goes to 1/-2. i hope that helped!

4 0

2 years ago

Read 2 more answers

2,7,26,101,400 what is the next number

Sedaia [141]

$a_1=2\\a_2=4*a_1-1=4*2-1=7\\a_3=4*a_2-2=4*7-2=26\\a_4=4*a_3-3=4*26-3=101\\a_5=4*a_4-4=4*101-4=400\\a_6=4*a_5-5=4*400-5=1595$

4 0

3 years ago