Ask question

Ask question

All categories

2 years ago

7

The question is "the polynomial is even & positive, does it have 3 or 4 zeros?

2 answers:

Pavel [41]2 years ago

5 0

Answer:

1

Step-by-step explanation:

sergejj [24]2 years ago

4 0

Even function is symmetrical about the y axis
not sure what it means if the polynomial is positive

the zeroes are how many times it hits the x axis
it hits 4 times so 4 zeros

You might be interested in

Use the Order of Operations (O.O.O) to solve each expression. <br><br> (10-5) 2· 5)- 2 2

sattari [20]

5•2•5)-2 2
10•5)-2 2
50-2 2
48•2
96
So the answer is 96

7 0

2 years ago

Margo is designing a label. The dimensions of the label are 3 1/2 inches by 1 1/4 inches. What is the area of the label?

Damm [24]

Answer:

4.375 sq. in.

Step-by-step explanation:

multiply 1.25 by 3.5 and you get 4.375

4 0

2 years ago

Find the area of the small rectangle. 2x + 3 x + 6 x - 5

Sloan [31]

Answer:

11x-5

Step-by-step explanation:

8 0

2 years ago

Find two vectors in R2 with Euclidian Norm 1<br> 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

2 years ago

A fisherman sold some big fish at $4 each and twice as many small fish at

harina [27]

Answer:

The answer is 12 big fish.

5 0

2 years ago