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Sergio039 [100]
3 years ago
15

Which could be the function graphed below?

Mathematics
1 answer:
Lina20 [59]3 years ago
3 0

Answer:A

Step-by-step explanation:

if you look at it carefully the endpoint is at the bottom of the Y value which looks like negative 2 this is the only answer you can be

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Complex Numbers
seropon [69]

Answer:

±10

Step-by-step explanation:

sqrt(-4) * sqrt(-25).

We know that the sqrt(a) sqrt(b) = sqrt(ab)

sqrt(-4*-25)

sqrt(100)

±10

4 0
3 years ago
Carina's car averages 25 miles per gallon.Will she be able to make the round trip on one tank of gas.
ser-zykov [4K]
There is not enough information to answer the question. To answer it, you need to know how much gas is in her gas tank and how far she needs to travel.
4 0
3 years ago
Help please thank you:) QUICKLY!!!!!!!!!
STALIN [3.7K]

Answer:

y = -8

Step-by-step explanation:

2x2 - 12

4 - 12 = -8

hope this helps

4 0
3 years ago
Read 2 more answers
All points of the step function f(x) are graphed.
ValentinkaMS [17]

Answer:

i think its 2 my friend

Step-by-step explanation:

8 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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