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3 years ago

List all the possible rational zeros of p(x)=3x^3-3x^2+4x-12

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

3 0

Answer:

$x = 3$

$x)=x^3-3x^2-4x+12 \\ 0 = 3x ^2 - 3x ^2 + 4x - 12 \\ 0 = 4x - 12 \\ - 4x = - 12 \\ - x = - 12 \div 4 \\ x = 3$

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Find two vectors in R2 with Euclidian Norm 1 whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .

Remembering,

$y_1=-3x_1\\y_2=-3x_2$

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

5 0

3 years ago

Determine if line KM and line ST are ॥, ⊥, or neither: K(-5,1) , M(-2,-8), and S(12,14), T(20,-10) Parallel Perpendicular or nei

Black_prince [1.1K]

Answer:

The lines are parallel i.e. KM || ST

Step-by-step explanation:

Given

K(-5,1) , M(-2,-8), and S(12,14), T(20,-10)

In order to find if the lines are parallel, perpendicular or neither slope will be used.

The slope of any line which passes from two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula

$m = \frac{y_2-y_1}{x_2-x_1}$

If two lines are parallel their slope is equal. If two lines are perpendicular the product of their slope is equal to -1

So, first slopes of both lines will be calculated.

Let m1 be the slope of KM

and

m2 be the slope of ST

In case of KM:

$(x_1,y_1) = (-5,1)\\(x_2,y_2) = (-2,-8)$

Putting the values in the formula for slope

$m_1 = \frac{-8-1}{-2-(-5)}\\m_1 = \frac{-9}{-2+5}\\= \frac{-9}{3}\\= -3$

In case of ST:

$(x_1,y_1) = (12,14)\\(x_2,y_2) = (20,-10)$

Putting the values in the formula for slope

$m_2 = \frac{-10-14}{20-12}\\= \frac{-24}{8}\\= -3$

As we can see that slope of both lines KM and ST are equal

i.e. m1 = m2 = -3

Hence,

The lines are parallel i.e. KM || ST

6 0

3 years ago

Please help me out . ill mark brain list

Korvikt [17]

Answer:

The amount of time Jill spent to do her homework

= 9 1/2 hours. which equals to 570 minutes.

Which means, the amount of times Jack needed to do his is

= 570 minutes x 1.5

= 855 minutes which equals to 14 1/4 hours

hope this helps

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3 years ago

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Which equation represents a linear function? Equation 1: y^3 = 2x +1 Equation 2: y= 3x +1 Equation 3: y= 5x^2 - 1 Equation 4: y=

Yuri [45]

the second equation is linear

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3 years ago

Solve-2a-5>3 Which graph shows the solution?

Ivanshal [37]

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2
a
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5
>
3
Step 1: Add 5 to both sides.
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5
+
5
>
3
+
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>
8
Step 2: Divide both sides by -2.
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<
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4
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6 0

3 years ago