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3 years ago

Find an equation for the nth term of a geometric sequence where the second and fifth terms are -8 and 512, respectively.

1 answer:

7 0

Easy
recall that
an=a1(r)^(n-1)
so

given 2nd and 5th term

we get
a2 and a5
so
a2=a1(r)^(2-1)=a1(r)^1=a1r
a5=a1(r)^(5-1)=a1(r)^4

also remember that $\frac{x^m}{x^n}=x^{m-n}$
so
$\frac{a_5}{a_2}= \frac{a_1r^4}{a_1r^1} =r^{4-1}=r^3= \frac{512}{-8}=-64$
so r^3=-64
cube root
r=-4
so

a2=a1r=-8
a2=a1(-4)=-8
divide both sides by -4
a1=2

so

equation is
$a_n=2(-4)^{n-1}$

C isi the answer

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Please assist me with the domain and range of graphs

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Answer: see below

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

Domain represents the x-values from the smallest (furthest left) to the biggest (furthest right).

Range represents the y-values from the lowest (furthest down) to the highest (furthest up).

Interval notation: If a value is included <em>(closed dot)</em>, use a bracket [ ]

If a value is NOT included <em>(open dot)</em>, use a parenthesis ( )

Note that ± ∞ is never included.

1. Domain: smallest x-value is -4 (included). biggest x-value is 3 (included)

Range: lowest y-value is 2 (included). highest y-value is 5 (included)

D: x = [-4, 3] R: y = [2, 5]

2. Domain: smallest x-value is -∞ (←). biggest x-value is 2 (included)

Range: lowest y-value is 2 (included). highest y-value is ∞ (↑)

D: x = (-∞, 2] R: y = [2, ∞)

3. Domain: smallest x-value is -∞ (←). biggest x-value is ∞ (→)

Range: lowest y-value is -∞ (↓). highest y-value is ∞ (↑)

D: x = (-∞, ∞) R: y = (-∞, ∞)

4. Domain: smallest x-value is -∞ (←). biggest x-value is ∞ (→)

Range: lowest y-value is -∞ (↓). highest y-value is 4 (included)

D: x = (-∞, ∞) R: y = (-∞, 4]

5. Domain: smallest x-value is -∞ (←). biggest x-value is ∞ (→)

Range: lowest y-value is 3 (included). highest y-value is ∞ (↑)

D: x = (-∞, ∞) R: y = [3, ∞)

6. Domain: smallest x-value is -2 (included). biggest x-value is ∞ (→)

Range: lowest y-value is -3 (included). highest y-value is ∞ (↑)

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3 years ago

Write an equation With the slope equal to -4/5 and (0,8)

Murljashka [212]

Answer: y = (-4/5)x + 8

Solution

The equation of line is

y = mx + b

Where

y = y coordinate

m = slope

x = x coordinate

b = y intercept

In order to write the equation of line we need to find the y-intercept (b)

substitute the slope (-4/5) and coordinates given (0,8) ,

y = mx + b

8 = -4/5 (0) + b

8 = b

So,

y = (-4/5)x + 8

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3 years ago

If g(x)=4x^2-16 we’re shifted 9 united to the right and 1 down, what would the new equation be?

Inessa05 [86]

Shown in the graph

<h2>Explanation:</h2>

Using graph tools we can graph the function:

$g(x)=4x^2-16$

which is the red graph shown below. As you can see, this is a parabola. The rule for vertical and horizontal shifts is as follows:

$Let \ c \ be \ a \ positive \ real \ number. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:$

$\bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ h(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ h(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ right \ \mathbf{right}: \\ h(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ left \ \mathbf{left}: \\ h(x)=f(x+c)$

Therefore, If we shift the red graph 9 units to the right and 1 down, our new function (let's call it $h(x)$ ) will be:

$h(x)=4\left(x-9\right)^{2}-16-1 \\ \\ Simplifying: \\ \\ h(x)=4\left(x-9\right)^{2}-17$

This graph is the blue graph below. Let's verify the transformation taking the vertex of the red graph:

$(0,-16)$

By translating the 9 units to the right and 1 down the vertex is also translated by the same rule, so:

$New \ vertex: \\ \\ (0+9,-16-1) \\ \\ (9,-17)$

<h2>Learn more:</h2>

Cubic function: brainly.com/question/13773618#

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