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Umnica [9.8K]
3 years ago
12

What is the derivative of x^2 - x + 3 at the point x = 5?​

Physics
1 answer:
Hunter-Best [27]3 years ago
5 0

Answer:

9

Explanation:

d/dx (x² - x + 3)

= 2x - 1

when x = 5,

2x-1

= 2(5) - 1

= 10 - 1

= 9

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What is the new pressure of 150 ml of a gas that is compressed to 50 ml when the original pressure was 3.0 atm and the temperatu
Firlakuza [10]

Answer: 9.0 atm

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=3.0atm\\V_1=150mL\\P_2=?mmHg\\V_2=50mL

Putting values in above equation, we get:

3.0\times 150mL=P_2\times 50mL\\\\P_2=9.0atm

Thus new pressure of 150 ml of a gas that is compressed to 50 ml is 9.0 atm

5 0
4 years ago
Please need help on this
Andrei [34K]

Q1 option 3

I hope it helps

3 0
3 years ago
Can molecules with double or triple bonds twist
stiks02 [169]

Answer:

No.

Explanation:

The only way a twist may be done is if the trans form of an alkene/alkyne is twisted into the cis form--only if/when the pi bond is brokwn.

6 0
3 years ago
Which two actions best demonstrate taking responsibility?
kow [346]

Answer:

Leading your people into the right direction, and always knowing what is best for  you and your people.

Explanation:

6 0
4 years ago
Suppose a small quantity of radon gas, which has a half-life of 3.8 days, is accidentally released into the air in a laboratory.
Daniel [21]

Answer:

1 day

Explanation:

Let the safe level = x

The current level = x + 0.2x = 1.2 x

Thus,

Half life = 3.8 days

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{3.8}\ days^{-1}

The rate constant, k = 0.1824 days⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

So,

\frac {[A_t]}{[A_0]} = x / 1.2 x = 0.8333

t = ?

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.8333=e^{-0.1824\times t}

t ≅ 1 day

<u>Lab must be vacated in 1 day.</u>

4 0
3 years ago
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