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2 years ago

8

I need help someone help asap please I'm late I have been needing help thanks in advance and I appreciate it

1 answer:

Nadusha1986 [10]2 years ago

6 0

Answer: I believe it would be tan(-5pi/6)

Step-by-step explanation:

Since tan is he length of the side opposite the angle divided by the length of the adjacent side.

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Tìm mệnh đề đúng trong các mệnh đề sau:

timurjin [86]

Answer:

What do u mean?

Step-by-step explanation:

3 0

2 years ago

The petrol tank at the local service station has a capacity of 6000L. It is 66.06 percent full. How many litres of petrol are ne

Rina8888 [55]

Since the tank is 66.06% full, it is (100 - 66.06) = 33.94% empty.

Its capacity is 6,000L. So, in order to fill it to capacity, the service station
must order 33.94% of 6,000L .

33.94% = 0.3394
"of" means "times"

0.3394 x 6,000L = <em>2,036.4 liters</em> needed, to fill the tank to the rim.

7 0

3 years ago

(EASY)(BRAINLIEST)Which of the four angles marked on the picnic table is obtuse?

Nataly [62]

The angle that is obtuse is 1, because it is greater than 90°.

Hope this helps and have a nice day:)

4 0

2 years ago

Read 2 more answers

lesya [120]

1_ 0/4. ( if toss a coin twice, we won't have three results)

2_ 0/4

3_ 3/4

4_ 3/4

5 _ 1/4

6_ 1/4

good luck

4 0

3 years ago

The rate of growth of profit (in millions) from an invention is approximated by Upper P prime (x )equals x e Superscript negati

Alecsey [184]

Answer:

The function is $P(x) = \frac{1}{2} e^{-x^2} +0.016$

Step-by-step explanation:

From the question we are told that

The rate of growth is $P'(x) = xe^{x} - x^2$

The total profit is $P(2) =$ $25,000

The time taken to make the profit is $x = 2 \ years$

From the question

$P'(x) = xe^{-x^2}$ is the rate of growth

Now here x represent the time taken

Now the total profit is mathematically represented as

$P(x) = \int\limits {P'(x)} \, = \int\limits {xe^{-x^2}} \,$

So using substitution method

We have that

$u = - x^2$

$du = 2xdx$

So

$p(x) = \int\limits {\frac{1}{2} e^{-u}} \, du$

$p(x) = {\frac{1}{2} [ e^{-u}} +c ]$

$p(x) = {\frac{1}{2} e^{-x^2}} + \frac{1}{2} c$ recall $u = - x^2$ and let $\frac{1}{2} c = Z$

At x = 2 years

$P(x) =$ $25,000

So

Since the profit rate is in million

$P(x) =$ $25,000 = $\frac{25000}{1000000} =$ $0.025 millon dollars

So

$0.025 = {\frac{1}{2} e^{-2^2}} + Z$

=> $Z = 0.025 - {\frac{1}{2} e^{-2^2}}$

$Z = 0.016$

So the profit function becomes

$P(x) = \frac{1}{2} e^{-x^2} +0.016$

5 0

2 years ago