Find the next two terms of the following sequence 13, 4, -11,- 32,...

1 answer:

4 0

13 - 9 = 4, 4 - 15 = -11, -11 - 21 = -32
What you're subtracting by increases by 6 each time.

The next two terms are -59 and -92; if you need me to explain this answer further, let me know.

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2. There are 1500 candies in a candy dish. Each day half of the candy is taken out.

katen-ka-za [31]

Answer:

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Step-by-step explanation:

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4 years ago

CHECK MY ANSWER!? Which of the following is true about the polynomial 3x^6-4? Select all that apply.

Ivenika [448]

The polynomial is written in standard form and has three terms.

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3 years ago

Use the Rational Zero Theorem to list all possible rational zeros for the given function. f(x) = 10x^5 + 8x^4 - 15x^3 +2x^2 - 2

vagabundo [1.1K]

Given:

The function is:

$f(x)=10x^5+8x^4-15x^3+2x^2-2$

To find:

All the possible rational zeros for the given function by using the Rational Zero Theorem.

Solution:

According to the rational root theorem, all the rational roots are of the form $\dfrac{p}{q},q\neq 0$ , where p is a factor of constant term and q is a factor of leading coefficient.

We have,

$f(x)=10x^5+8x^4-15x^3+2x^2-2$

Here,

Constant term = -2

Leading coefficient = 10

Factors of -2 are ±1, ±2.

Factors of 10 are ±1, ±2, ±5, ±10.

Using the rational root theorem, all the possible rational roots are:

$x=\pm 1,\pm 2,\pm \dfrac{1}{2}, \pm \dfrac{1}{5},\pm \dfrac{2}{5},\pm \dfrac{1}{10}$ .

Therefore, all the possible rational roots of the given function are $\pm 1,\pm 2,\pm \dfrac{1}{2}, \pm \dfrac{1}{5},\pm \dfrac{2}{5},\pm \dfrac{1}{10}$ .

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3 years ago

Find the standard form of the equation of the parabola with a focus at (-2, 0) and a directrix at x = 2.

krok68 [10]

Answer:

$y^2=\frac{1}{8}x$

Step-by-step explanation:

The focus lies on the x axis and the directrix is a vertical line through x = 2. The parabola, by nature, wraps around the focus, or "forms" its shape about the focus. That means that this is a "sideways" parabola, a "y^2" type instead of an "x^2" type. The standard form for this type is

$(x-h)=4p(y-k)^2$

where h and k are the coordinates of the vertex and p is the distance from the vertex to either the focus or the directrix (that distance is the same; we only need to find one). That means that the vertex has to be equidistant from the focus and the directrix. If the focus is at x = -2 y = 0 and the directrix is at x = 2, midway between them is the origin (0, 0). So h = 0 and k = 0. p is the number of units from the vertex to the focus (or directrix). That means that p=2. We fill in our equation now with the info we have:

$(x-0)=4(2)(y-0)^2$