All categories

3 years ago

Hey here some points free but add me as a friend bc ya know

Mathematics

2 answers:

jek_recluse [69]3 years ago

6 0

Answer:

added!

Step-by-step explanation:

just olya [345]3 years ago

5 0

Answer:

yes they are

Step-by-step explanation:

my last brain cell hurt

You might be interested in

So I'm still trying to figure out but tell me. What are two demicals rounded to 1.5

Andrej [43]

451.76 and <span>451.7576 these two</span>

7 0

3 years ago

Please help this is Science.

monitta

C. velocity includes the direction of travel whereas speed does not.

6 0

2 years ago

Read 2 more answers

I need the answer for the question -4(x-10)+8 for my homework please.

mixer [17]

Answer:

i think the answer is

-4x+48

7 0

3 years ago

Simplify (1 − cos x)(1 + cos x). (2 points)

Vadim26 [7]

Answer:

sin²x

Step-by-step explanation:

we have

(1 − cos x)(1 + cos x)=1-cos²x

Remember that

sin²x+cos²x=1 ------> i-cos²x=sin²x

therefore

(1 − cos x)(1 + cos x)=sin²x

4 0

3 years ago

Read 2 more answers

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

3 years ago