All categories

2 years ago

Factor completely

Mathematics

2 answers:

boyakko [2]2 years ago

8 0

Answer:

hope this helps:))))))

Orlov [11]2 years ago

5 0

Answer:

(x-3)(x+3)(x+2)

Step-by-step explanation:

x^3 + 2x^2 - 9x -18

= x^2(x+2) - 9(x+2)

= (x^2 - 9) (x+2)

= (x+3)(x-3)(x+2)

You might be interested in

Please help with this will give BRAINLIST to best answer

Aleonysh [2.5K]

Answer:

pxp^8=12

Step-by-step explanation:

7 0

3 years ago

Denote the set below by set-builder notation, using x as the variable.

Temka [501]

Answer:

C. {x|x is an odd natural number between 1 and 67}

Step-by-step explanation:

B and C is also correct if the natural number is defined to be started with 1: positive integer,

but if it's defined to be non-negative real number then only C

7 0

3 years ago

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 $x_2$ .