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ELEN [110]
3 years ago
6

Which sequences are geometric? Check all that apply.

Mathematics
1 answer:
Sliva [168]3 years ago
4 0

Answer:

1, 5, 25, 125, ...

3, 6, 12, 24, ...

Step-by-step explanation:

  • <em>a geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio</em>

1, 5, 25, 125, ...

  • yes, the common ratio is 5

3, 6, 9, 12,...

  • no

3, 6, 12, 24, ...

  • yes, the common ratio is 2

3, 9, 81, 6, 561, ...

  • no

10, 20, 40, 60, ...

  • no
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How do you do this question?
vaieri [72.5K]

Answer:

(-∞, -2), (-2, -0.618), and (1.618, 3)

Step-by-step explanation:

The red plus (+) signs indicate the regions in which the function is concave up, and the red negative (-) signs indicate the regions in which the function is concave down.

Note that the sign of the concavity changes at an inflection point.

Let's examine the intervals given.

(-∞, -2): Yes, concave up.

(-∞, -1.17): No. Concave up in (-∞, -2) but concave down in (-2, -1.17).

(-2, 0): No. Concave down in (-2, -1.17) but increasing in (-1.17, 0.0).

(-1.17, 0.689): Yes. Concave up.

(-0.618, 1.618): No. Concave up in (-0.618, 0.689) but concave down in (0.689, 1.618).

(0, 3): No. Concave up in (0, 0.689), concave down in (0.689, 2.481), and concave up in (2.481, 3).

(2.481, ∞): Yes. Concave up.

The three intervals that are concave up are (-∞, -2), (-1.17, -0.689), and (2.481, ∞).

4 0
3 years ago
Find the equation of a parabola with a vertical axis and its vertex at the origin and passing through the point (-2, 3)
vredina [299]

a vertical axis, I assume it means a vertical axis of symmetry, thus it'd be a vertical parabola, like the one in the picture below.

\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \qquad \leftarrow vertical\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=0 \end{cases}\implies y=a(x-0)^2+0 \\\\\\ \textit{we also know that } \begin{cases} x=-2\\ y=3 \end{cases}\implies 3=a(-2-0)^2+0\implies 3=4a \\\\\\ \cfrac{3}{4}=a~\hspace{10em}y=\cfrac{3}{4}(x-0)^2+0\implies \boxed{y=\cfrac{3}{4}x^2}

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
Write the point-slope form of the equation of the line that passes through the points (6, -9) and (7, 1). Include your work in y
lys-0071 [83]
First find the slope, 
m=(y2-y1)/(x2-x1)=(1-(-9))/(7-6)=10
Next, use the point slope form to find the equation:
L : y-y1=m(x-x1) => y-1=10(x-7) 
    [in point slope form, point is (7,1), slope=m=10 ]


5 0
3 years ago
PLEASE HURRY!!
barxatty [35]

Answer: Is a function

the question unfinished so it will be hard to solve

3 0
3 years ago
Read 2 more answers
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