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1 year ago

8

Can someone help me please?? It’s due on thursday ❤️ (rotate picture if u can)

1 answer:

rusak2 [61]1 year ago

7 0

Answer: The two companies never charge the same amount to rent a bicycle for the same number of hours.

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1/3 + a = 5/4<br> A=? ..........................................................

NARA [144]

11/12 I’m pretty sure

3 0

2 years ago

Which number is a perfect cube?<br> O9<br> O 18<br> 0 27<br> O 36

Arturiano [62]

Answer:

27

Step-by-step explanation:

9 is not a perfect cube.

18 is not a perfect cube.

27 is a perfect cube.

∛27 = 3

36 is not a perfect cube.

8 0

2 years ago

Read 2 more answers

of the 273 students in a school, 95 volunteered to work the book sale. About what percent of the students volunteered?

Whitepunk [10]

28.7% to b exact
I hope this helped you

6 0

3 years ago

Which lines best approximate the directrices of the ellipse? Round to the nearest tenth.

Mrrafil [7]

C aka y = -4.6 and y = 4.6

Step-by-step explanation:

i just took the unit review test on edg. :)

6 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