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2 years ago

13

Mario needs 1024 MB to make a backup of his computer files, for this he has 3 USB memories with a capacity of 256 MB. How many U

SB sticks with a 128 MB capacity does Mario need for this task?

2 answers:

Brut [27]2 years ago

6 0

256 x 3 is uhh 768 so it would be 2 of the 128 sticks

grigory [225]2 years ago

4 0

………. i don’t even know if that makes sense….

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Plot -1, 8, and -11 on the number line.<br> 5<br> 10<br> -10<br> -5<br> 0<br> X

svetlana [45]

-1 is the first line after 0

8 is 3rd line one the right side after 5

-11 is just first line after 10..

I hope U understand

8 0

3 years ago

Read 2 more answers

A company manufacturers video games with a current defect rate of 0.95%. To make sure as few defective video games are delivered

RSB [31]

There would be about 19 defective products delivered.

First, let's start with the number of defective games.
100000 x 0.0095 = 950

Now, the test will catch 98% of those defects. That means 2% of the defects will get through to the consumers.
0.02 x 950 = 19

3 0

3 years ago

Read 2 more answers

Find the Range and Domain of the Binary relations

sukhopar [10]

Answer:

A = {3,5,6}

B = {1,3}

<u><em>A x B</em></u> = {(3,1),(3,3),(5,1),(5,3),(6,1),(6,3)}

Domain of A x B = (The x figures in ascending order)

=> {3,5,6} (Figures repeating more than once must be written only 1 time)

Range of A x B = (The y figures in ascending order)

=> {1,3}

<u><em>B x A </em></u>= {(1,3),(3,3),(1,5),(3,5),(1,6),(3,6)}

Domain of B x A = {1,3}

Range of A x B = {3,5,6}

8 0

3 years ago

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is:

$(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})$

$(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]$

$(x,y) = (10.452, 34.531)\,[km]$

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

$\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}$

$\theta \approx 16.840^{\circ}$

And the distance from the camp is calculated by the Pythagorean Theorem:

$r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}$

$r = 36.078\,km$

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

5 0

3 years ago

HELP WITH THESE QUESTIONS PART 2... 1 Question

Lunna [17]

<u>Step-by-step explanation</u> and Answers:

$(\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}})$ is in the format (cos, sin). So, adjacent = 1, opposite = 3, hypotenuse = $\sqrt{10}$

sin = $\frac{opposite}{hypotenuse}$ = $\frac{3}{\sqrt{10}}(\frac{\sqrt{10}}{\sqrt{10}})$ = $\frac{3\sqrt{10}}{10}$

cos = $\frac{adjacent}{hypotenuse}$ = $\frac{1}{\sqrt{10}}(\frac{\sqrt{10}}{\sqrt{10}})$ = $\frac{\sqrt{10}}{10}$

tan = $\frac{opposite}{adjacent}$ = $\frac{3}{1}$ = 3

csc = $\frac{hypotenuse}{opposite}$ = $\frac{\sqrt{10}}{3}$

sec = $\frac{hypotenuse}{adjacent}$ = $\frac{\sqrt{10}}{1}$ = $\sqrt{10}$

cot = $\frac{adjacent}{opposite}$ = $\frac{1}{3}$

7 0

3 years ago