Ask question

Ask question

All categories

3 years ago

8

Morty buys and sells computer parts. He bought 2 monitors for $25.00 each and later sold them for $88.00 each. He bought 4 cases

for $15.00 each and later sold them for $24.00 each. He bought 5 memory modules for $20.00 each and later sold them for $55.00 each

2 answers:

masya89 [10]3 years ago

8 0

What’s the question tho

n200080 [17]3 years ago

3 0

What is the question

You might be interested in

A submarine changes its position by −7 meters every second.

IgorLugansk [536]

The correct anwer is -21 and how i got that is by taking -7*300 to get -21 as my answer let me know if it helped any...

5 0

2 years ago

Read 2 more answers

Trevor is reading a book for a book report. Last week, he read 35 pages of his book. This week he read 2.5 times as many pages a

Svet_ta [14]

Answer:85.5

Step-by-step explanation:

Use calculator

3 0

3 years ago

I need help on this problem someone please help

Hatshy [7]

The outlier is a because it is the most different value compared to the rest.

5 0

3 years ago

Read 2 more answers

Sboob wohs dna ten.esaelpemetaD ot og FG evah i naC

emmainna [20.7K]

Answer:

Ummmm no thanks. I am not going there.

Step-by-step explanation:

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