Ask question

Ask question

All categories

3 years ago

8

Help me out pls thanks!

1 answer:

natulia [17]3 years ago

6 0

The answer is 2 I hope this is right

You might be interested in

HELL PLEASE ASAP 25 POINTS PLEASE ASAPPPP

Andru [333]

Well I think it’s the third one but this is really advanced

7 0

3 years ago

Help please! I’ll give brainliest

NeTakaya

1)y=3x
2)y=x
3)y=7x
4)y=x+7
4)y=2x+1
6)y=12x+2

5 0

3 years ago

Please help me ;v; its hard-

Fed [463]

Ask a teacher? :)!!!!!!!!!!!!

7 0

3 years ago

Read 2 more answers

Describe how a company that produce earphones can design a survey

DiKsa [7]

First of all this survey need to be <span>anonymous. The company need to know the population whom the survey is addressed. So, I think the following questions can be a good example:

1. </span><span>Where do you usually use earphones? (Please select all that apply) 
</span><span>library
home
transportation
kitchen
driving/riding
Other

2. What are the purposes you use earphones? (Please select all that apply)</span><span> learn language
listen to music
watch video
privacy concerns(show ‘don’t bother me’)
for beauty
Other
</span>
<span>3. What kind of earphones you like the most? <span>
circumaural
supra-aural
earbuds
in-ear

<span>4. What are the reasons make you like the earphone you chose in question 3? (Please select all that apply) <span>
appearance
comfortability
function(plays music well)
brand
price
portability
Other

<span>5. Have you ever felt vulnerable or unsafe while wearing headphones and listening to music because you are not as aware of your surroundings? <span><span>
yes</span><span>
no</span></span></span></span></span></span></span>

3 0

3 years ago

: Show that the solution of the differential equation: = − − − − − is of the form: + + ( − ) = + , When = and =

Serhud [2]

Answer:

$y = \tan(x + \frac{x^2}{2})$

Step-by-step explanation:

Poorly formatted question; The complete question requires that we prove that $y=\tan(x+\frac{x\²}{2})$

When

$\frac{dy}{dx} =1+xy\²+x+y\²$ and $y(0)=0$

We have:

$\frac{dy}{dx} =1+xy\²+x+y\²$

Rewrite as:

$\frac{dy}{dx} =1+x+xy\²+y\²$

Factorize

$\frac{dy}{dx} = (1+x)+y\²(x+1)$

Rewrite as:

$\frac{dy}{dx} = (1+x)+y\²(1+x)$

Factor out 1 + x

$\frac{dy}{dx} = (1+y\²)(1+x)$

Multiply both sides by $\frac{dx}{1 + y^2}$

$\frac{dy}{1+y\²} = (1+x)dx$

Integrate both sides

$\int \frac{dy}{1+y\²} = \int (1+x)dx$

Rewrite as:

$\int \frac{1}{1+y\²} dy = \int (1+x)dx$

Integrate the left-hand side

$\int \frac{1}{1+y\²} dy = \tan^{-1}y$

Integrate the right-hand side

$\tan^{-1}y = x + \frac{x^2}{2} + c$

$y(0)=0$ implies that: $(x,y) = (0,0)$

So:

$\tan^{-1}y = x + \frac{x^2}{2} + c$ becomes

$\tan^{-1}(0) = 0 + \frac{0^2}{2} + c$

This gives:

$0 = 0 +0 + c$

$0 =c$

$c = 0$

The equation $\tan^{-1}y = x + \frac{x^2}{2} + c$ becomes

$\tan^{-1}y = x + \frac{x^2}{2} + 0$

$\tan^{-1}y = x + \frac{x^2}{2}$

Take tan of both sides

$y = \tan(x + \frac{x^2}{2})$ --- Proved

8 0

3 years ago