I need help I don’t know how to do this

Plz help me fastttt it surface area of a rectangular prism

horrorfan [7]

Answer:

72.2 square meter

Step-by-step explanation:

(3.8)(3.8)= 14.44

(14.44)(5)=72.2

4 0

3 years ago

Pls help...also ill mark brainlest

pentagon [3]

Answer:

D

Step-by-step explanation:

I think it is D

4 0

2 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

4 years ago

Look at the table representing the distance covered by a car accelerating from 0 miles per hour. The table shows a relationship.

vfiekz [6]

quadratic

second

16

8

✌️

8 0

3 years ago

300 ml of pure alcohol is poured from a bottle containing 2 l of pure alcohol. Then, 300 ml of water is added into the bottle. A

Ede4ka [16]

Answer:

The present percentage of pure alcohol in the solution is 72.25% of pure alcohol

Step-by-step explanation:

The volume of pure alcohol poured from the 2 l bottle of pure alcohol = 300 ml of pure alcohol

The volume of water added into the bottle after pouring out the pure alcohol = 300 ml of water

The volume of diluted alcohol poured out of the bottle = 300 ml of diluted alcohol

The volume of water added into the bottle of diluted alcohol after pouring out the 300 ml of diluted alcohol = 300 ml of water

Step 1

After pouring the 300 ml of pure alcohol and adding 300 ml of water to the bottle, the percentage concentration, C%₁ is given as follows;

C%₁ = (Volume of pure alcohol)/(Total volume of the solution) × 100

The volume of pure alcohol in the bottle = 2 l - 300 ml = 1,700 ml

The total volume of the solution = The volume of pure alcohol in the bottle + The volume of water added = 1,700 ml + 300 ml = 2,000 ml = 2 l

∴ C%₁ = (1,700 ml)/(2,000 ml) × 100 = 85% percent alcohol

Step 2

After pouring out 300 ml diluted alcohol from the 2,000 ml, 85% alcohol and adding 300 ml of water, we have;

Volume of 85% alcohol = 2,000 ml - 300 ml = 1,700 ml

The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol = 85/100 × 1,700 = 1,445 ml

The total volume of the diluted solution = The volume of the 85% alcohol in the solution + The volume of water added

∴ The total volume of the twice diluted solution = 1,700 ml + 300 ml = 2,000 ml

The present percentage of pure alcohol in the solution, C%₂ = (The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol)/(The total volume of the diluted solution) × 100

∴ C%₂ = (1,445 ml)/(2,000 ml) × 100 = 72.25 %

The present percentage of pure alcohol in the solution, C%₂ = 72.25%

3 0

3 years ago