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Aleonysh [2.5K]

3 years ago

12

solve for a and B. A. a=400/21, b = 580/21 B. a=400/21, b=20/21. C a=580/21, b=20/21. D a=20/21, b=580/21

1 answer:

Brut [27]3 years ago

5 0

Answer:

Step-by-step explanation:

21/29 = 20/b

b = 20*29/21 = 580/21

a^2 + 20^2 = (580/21)^2

a = 400/21

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What is the solution to the equation?

AleksandrR [38]

Solution of the equation: $x=6\frac{1}{2}$

Step-by-step explanation:

The equation that we have to solve in this problem is:

$13\frac{3}{4}+x=7\frac{1}{4}$

The first step to do is to rewrite the mixed fractions as improper fractions. We have:

$13\frac{3}{4}=\frac{13\cdot 4+3}{4}=\frac{52+3}{4}=\frac{55}{4}$

And

$7\frac{1}{4}=\frac{7\cdot 4+1}{4}=\frac{29}{4}$

So the equation becomes

$\frac{29}{4}+x=\frac{55}{4}$

Now we multiply by 4 each term on both sides, and we get

$29+4x=55$

Now we subtract 29 from both sides,

$4x=55-29=26$

And finally, we divide both sides by 4:

$x=\frac{26}{4}=\frac{13}{2}=6\frac{1}{2}$

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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:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

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2 years ago