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2 years ago

Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function. f(x) = -2x4 + 4x3 + 3x2 + 18

1 answer:

Nookie1986 [14]2 years ago

6 0

$f(x) = -2x^4 + 4x^3 + 3x^2 + 18$

$18:\{\pm1;\ \pm2;\ \pm3;\ \pm6;\ \pm9;\ \pm18\}\\2:\{\pm1;\ \pm2\}\\\\Answer:\boxed{\{\pm\frac{1}{2};\ \pm1;\ \pm\frac{3}{2};\ \pm2;\ \pm3;\ \pm\frac{9}{2};\ \pm6;\ \pm9;\ \pm18\}}$

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H + 169 = 759 solve for h

steposvetlana [31]

Answer:

h = 590

Step-by-step explanation:

To find x, rearrange the equation to where h is on its own side.

<u>Rearranged equation:</u>

$759 - 169 = h$

Since 169 added to h was to find 769, 169 subtracted from 759 represents h.

<u>Solve:</u>

$759 - 169 = 590$

590 = h.

5 0

2 years ago

one number is 4 more than another number. if 4 times the smaller number is decreased by twice the larger number. the result is 1

soldier1979 [14.2K]

x and x+4 are the numbers

3x=2(x+4)

3x=2x+8

3x-2x=8

x=8

x+4=12

3 0

2 years ago

A cook needs 4 cups of vegetable broth for a soup recipe. How much is this in pints?

kramer

I believe the answer is 2 because 1/2 x 4 = 4/2 =2

7 0

2 years ago

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

7 months ago

Help me please i will mark brainliest if correct

iren2701 [21]

Answer:

7 x 2 + 31

Step-by-step explanation:

Equals to the bottom line answer is 7

7 0

2 years ago