Divide and write in simplest form 3/10 divided by 1 2/3

greg s homework assignment was to find allthe common factors of 28 and 42 so greg made a list of all the factors of 28 and 42 he

kicyunya [14]

The common factors for 28 and 42 are 1, 2, 7 and 14.

3 0

3 years ago

3 1 ) How do -1- and -1- compare? Choose a symbol to make the statement true. 4 4 3 -1- ? 1 4 4

lukranit [14]

Answer:

4 or-1- i dont know

Step-by-step explanation:

5 0

3 years ago

ANSWER FAST PLZZZ Which polygons are congruent? Select each correct answer. Two right scalene triangles labeled D E F and P Q R.

Shalnov [3]

A is congruent
B is congruent
C is congruent
D is not congruent
And E is the same picture as B so it is also congruent

7 0

3 years ago

Read 2 more answers

There are eight shirts in your closet, three blue, two are black, one is green, and two are white. You randomly select one to we

Maru [420]

The probability of wearing blue shirts on both days is 3/28.

<h3>What is probability?</h3>

Probability is the measure of the likelihood of an occurrence.

Given:

8 shirts are available out of which 3 are blue, 1 is green and 2 are white.

P(blue on Monday) = 3/8

Now, 7 shirts are left to wear on Tuesday

P(blue on Tuesday) =2/7

So,

the probability of wearing blue shirts on both days = 3/8 * 2/7 = 3/28

Learn more about probability here:

brainly.com/question/24756209

#SPJ1

4 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

3 years ago