Ask question

Ask question

All categories

2 years ago

11

Pls help i’ll mark brainliest!!

1 answer:

Luba_88 [7]2 years ago

4 0

Answer: Aiden runs 187.5 meters per minute.
Explanation: Literally divide the two numbers

You might be interested in

What is the third quartile of the data set? 23, 35, 55, 61, 64, 67, 68, 71, 75, 94, 99

lara [203]

Answer:

75

Step-by-step explanation:

Q1=55 Q2=67 Q3=75

4 0

3 years ago

In the triangle shown, what is the value of the missing angle?

Contact [7]

Because angles in a triangle add up to 180, you can add up the given angles and then take that answer away from 180, so:
65 + 74 = 139
180 - 139 = 41°
So your answer is 41°, I hope this helps!

7 0

3 years ago

Read 2 more answers

. What is the value of x in the equation –6 + x = –2?

Usimov [2.4K]

4 is the right answer

5 0

3 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

3 years ago

Find the common difference of the arithmetic sequence -3, -9, -15

eduard

Answer:

they are subtracting 6 from each number

6 0

3 years ago