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3 years ago

([x/(x + 1)] - (4/5))/(x - 4)

Mathematics

1 answer:

Svetllana [295]3 years ago

3 0

Answer:

1/5x+5

Step-by-step explanation:

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CAN SOMEONE PLEASE HELP

Iteru [2.4K]

Here is how you would set it up...

You would to 12/100 = x/2 and cross multiply and solve by division. I believe. I'm almost positive that's how you do it. If that don't work out, comment and I'll respond.

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3 years ago

What is the formula of vat

balu736 [363]

Answer:

VAT = vat percent of selling price

Step-by-step explanation:

VAT = vat % × s.p

hope this answer will help you

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3 years ago

The standard normal curve shown here is a probability density curve for a continuous random variable. This means that the area u

faust18 [17]

the answer is 0.7068

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3 years ago

Complete parts a through c for the given function.

victus00 [196]

Answer:

a. The critcal points are at

$x=0,-5,3$

b. Then, $x = -5$ is a maximum and $x=3$ is a minimum

c. The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

Step-by-step explanation:

(a)

Remember that you need to find the points where

$f'(x)=0$

Therefore you have to solve this equation.

$20x^4 + 40x^3 - 300x^2 = 0$

From that equation you can factor out $20x^2$ and you would get

$20x^2 ( x^2 +2x - 15) = 0$

And from that you would have $20x^2 = 0$ , so $x = 0$ .

And you would also have $x^2 +2x-15 = 0$ .

You can factor that equation as $x^2 +2x -15 = (x+5)(x-3) = 0$

Therefore $x=-5 , x=3$ .

So the critcal points are at

$x=0,-5,3$

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

$f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\$

Then, $x = -5$ is a maximum and $x=3$ is a minimum

The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

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3 years ago

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An airplane pilot can see the top of a traffic control tower at a 20 degree angle of depression. the airplane is 5,000 feet away

daser333 [38]

The given question describes a right triangle with with one of the angles as 20 degrees and the side adjacent to the angle 20 degrees is of length 5,000 feet. We are looking for the length of the side opposite the angle 20 degrees.

Let the required length be x, then

$\tan{20^o}=\cfrac{opp}{hyp}=\cfrac{x}{5,000}\\ \\ \Rightarrow x=5,000\tan{20^o}=1,819.85$

Therefore, the height of the airplane above the tower is 1,819.85 feet.

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3 years ago

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