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2 years ago

PLEASE HELP

2 answers:

7 0

Osmosis is a special case of diffusion. Water, like other substances, moves from an area of higher concentration to one of lower concentration. ... In this example, the solute cannot diffuse through the membrane, but the water can. Water has a concentration gradient in this system.!!!!!!!!!!!!!!!!!!!!!:D!!!!!!!!!!!!!

xeze [42]2 years ago

5 0

Answer:

whatt is that the question

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Initially a ship and a submarine are stationary at sea level, positioned 1.78km apart. The submarine then manoeuvres to position

poizon [28]

I’m assuming ‘dives further’ means to go directly down

the angle of elevation of the ship from the submarine is equal to the angle of depression of the submarine from the ship, if we assume the sea level is perpendicular to ‘directly down’.

let both of these angles to be = $ when the submarine is at A and ¥ when the submarine is at B (excuse the lack of easily accessible variables as keys)

then this become a simple trig problem:

A)

Let O be the position of of the ship, and C be the original position of the submarine.

therefore, not considering direction

|OC| = 1.78km = 1780m

|CA| = 45m

these are the adjacent and opposite sides of a right angled triangle.

But tan($) = opp/adj = |CA|/|OC| = 45/1780

therefore $ = arctan(45/1780) which is roughly 1.45 degrees,

B)

similarly, noting that |CB| = |CA| + |AB| = 45 + 62 = 107m

tan(¥) = 107/1780

¥ = arctan(107/1780) which is roughly 3.44 degrees

4 0

2 years ago

600+600-200 Giving Brainliest

Mila [183]

Equation Answer is 1,000!

7 0

2 years ago

Read 2 more answers

Adam went on a shopping trip to Somerville. He purchased a set of dishes originally priced at

RUDIKE [14]

Answer:187.2

Step-by-step explanation:

Actual price $390

After 60% discount price will be 156

And Sales tax is 8% . Then the price of sales tax is $31.2.

156+31.2= $187.2

5 0

2 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago

H(x)=4|x+4|+3 what’s the answer

MatroZZZ [7]

Answer:

To find the transformation, compare the two functions and check to see if there is a horizontal or vertical shift, reflection about the x-axis, and if there is a vertical stretch.

Parent Function:

g(x)=|x|

Horizontal Shift: Right 4 Units

Vertical Shift: Up 1 Units

Reflection about the x-axis: None

Vertical Compression or Stretch: None

Step-by-step explanation:

4 0

2 years ago