All categories

2 years ago

11 A right triangle is shown.

Mathematics

1 answer:

Gre4nikov [31]2 years ago

3 0

Answer:

8 in.

Step-by-step explanation:

You might be interested in

Simplify: (64^-1/2)^-2/3

Ede4ka [16]

This = 64^( -1/2 * -2/3)

= 64 ^ 1/3

= cube root of 64

= 4 answer

5 0

3 years ago

The greatest common factor of 18 and 42 is: 02 6 09 18

Alex Ar [27]

Answer:

Step-by-step explanation:

We can obtain the greatest common factor by listing the factors of the 2 numbers, that is

Factors of 18 are 1, 2, 3, 6, 9, 18

Factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42

The common factors are 1, 2, 3, 6

The greatest common factor is 6

8 0

3 years ago

What is the solution to the inequality?

Pavel [41]

The steps to solving an inequality are: add or subtract from each side - multiply or divide both sides - simplify.

5x + 8 > -12
5x > -12 - 8
5x > -20
x > -20/5
x > -4

The answer is: x > -4

8 0

2 years ago

Read 2 more answers

Find two vectors in R2 with Euclidian Norm 1 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 $x_2$ .