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Novay_Z [31]
3 years ago
6

Guide

Mathematics
1 answer:
SIZIF [17.4K]3 years ago
5 0

Answer:

I need help with this as well.

Step-by-step explanation:

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I need some help, fast please!!
Sergio [31]

Answer:

6

Step-by-step explanation:

Integers are whole numbers that do not have decimals.

4.18 ÷ 0.68 = 6.14

Remember, round down with 0-4, round up with 5-9.

6.14 can be rounded down to 6.

3 0
3 years ago
Read 2 more answers
What is the slope of the line that passes through the pair of points? (4.7).(7,-2)
Tju [1.3M]

Answer:

-3      i hope this helps :)

Step-by-step explanation:

(4,7)    (7,-2)

x1 y1    x2 y2

<u>-2 - 7 </u> = <u>-9</u> = -3

7  - 4      3

5 0
3 years ago
Ms. Benjamin puts $1,200 in Virtue bank. The interest rate is 6% per year. a. How much interest will she get in after one year?
Mazyrski [523]

Answer:

the answer is 1

Step-by-step explanation:


3 0
3 years ago
Cole noticed that if he takes the opposite of his age and adds 40 he gets the number 28. how old is Cole?
AnnyKZ [126]
-x+40=28
-x=-12 (subtract 40 from both sides)

Divide by -1(in both sides)because you cant have a negative age.
COLE is 12 

 To check plug in the values.
5 0
3 years ago
A tank contains 100 L of water. A solution with a salt con- centration of 0.4 kg/L is added at a rate of 5 L/min. The solution i
Fantom [35]

Answer:

a) (dy/dt) = 2 - [3y/(100 + 2t)]

b) The solved differential equation gives

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

c) Concentration of salt in the tank after 20 minutes = 0.2275 kg/L

Step-by-step explanation:

First of, we take the overall balance for the system,

Let V = volume of solution in the tank at any time

The rate of change of the volume of solution in the tank = (Rate of flow into the tank) - (Rate of flow out of the tank)

The rate of change of the volume of solution = dV/dt

Rate of flow into the tank = Fᵢ = 5 L/min

Rate of flow out of the tank = F = 3 L/min

(dV/dt) = Fᵢ - F

(dV/dt) = (Fᵢ - F)

dV = (Fᵢ - F) dt

∫ dV = ∫ (Fᵢ - F) dt

Integrating the left hand side from 100 litres (initial volume) to V and the right hand side from 0 to t

V - 100 = (Fᵢ - F)t

V = 100 + (5 - 3)t

V = 100 + (2) t

V = (100 + 2t) L

Component balance for the amount of salt in the tank.

Let the initial amount of salt in the tank be y₀ = 0 kg

Let the rate of flow of the amount of salt coming into the tank = yᵢ = 0.4 kg/L × 5 L/min = 2 kg/min

Amount of salt in the tank, at any time = y kg

Concentration of salt in the tank at any time = (y/V) kg/L

Recall that V is the volume of water in the tank. V = 100 + 2t

Rate at which that amount of salt is leaving the tank = 3 L/min × (y/V) kg/L = (3y/V) kg/min

Rate of Change in the amount of salt in the tank = (Rate of flow of salt into the tank) - (Rate of flow of salt out of the tank)

(dy/dt) = 2 - (3y/V)

(dy/dt) = 2 - [3y/(100 + 2t)]

To solve this differential equation, it is done in the attached image to this question.

The solution of the differential equation is

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

c) Concentration after 20 minutes.

After 20 minutes, volume of water in tank will be

V(t) = 100 + 2t

V(20) = 100 + 2(20) = 140 L

Amount of salt in the tank after 20 minutes gives

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

y(20) = 0.4 [100 + 2(20)] - 40000 [100 + 2(20)]⁻¹•⁵

y(20) = 0.4 [100 + 40] - 40000 [100 + 40]⁻¹•⁵

y(20) = 0.4 [140] - 40000 [140]⁻¹•⁵

y(20) = 56 - 24.15 = 31.85 kg

Amount of salt in the tank after 20 minutes = 31.85 kg

Volume of water in the tank after 20 minutes = 140 L

Concentration of salt in the tank after 20 minutes = (31.85/140) = 0.2275 kg/L

Hope this Helps!!!

8 0
3 years ago
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