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Leto [7]
2 years ago
8

Point D is in the interior of ∠ABC. What is m∠DBC?

Mathematics
1 answer:
Rudiy272 years ago
3 0

37.5 degrees

we know angled like this always have 180 degrees total between the 2. which means we can set up a little equation to solve for x. The equation is (3x+22)+(x-4)= 180. Initially we can take the solid numbers (22 and -4) and get rid of them by taking them to the 180. 180- 22 is 158. 158 + 4 is 162. now we have 3x + x = 162. we combine like terms. means we end up with 4x =162. We now divide by 4 which means we get 40.5. Now we know x is 40.5. Now dbc is just 40.5- 4 which is 37.5.

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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
2 years ago
Jake is comparing the prices of two mattress cleaning companies. Company A charges $30 per mattress and an additional $13 as ser
Anna007 [38]
A)
y= total charges
x=number of mattresses
Company A: y=$30x+13 Company B: y=$28x+15

B)
plug x=4 into both equations to get
A: y=30(4)+13=$133
B: y=28(4)+15=$127
Company B charges less because 127<133

C)
Plug in 7 for x
A: y=30(7)+13=$223
B: y=28(7)+15=$211
A-B=223-211
=$12 Saved
6 0
2 years ago
Find the value of x​
andrew11 [14]
X=27

11x=16.5(18)
11x=297
x=27
8 0
3 years ago
What is 3/10 and 4/15 when they have a common denominator
NISA [10]

Answer:17/30

Step-by-step explanation:common denominator would be 30..

30÷10=3

30÷15=2

3x3=9

2x4=8

8+9=17

=17/30

3 0
3 years ago
given T(-2,4,7) and M(-3,5,2) find the ordered triple that represents TM Then find the magnitude of TM
kari74 [83]

Answer:

TM= (-6,20,14)

|TM|= 25.139 = 25.14

Step-by-step explanation:

Let's first calculate for TM

TM = dot product of The and M

TM= (-2,4,7) .(-3,5,2)

TM= (-2*-3),(4*5),(7*2)

TM= (-6,20,14)

Then for the magnitude of TM

|TM|= √(-6² + 20² + 14²)

|TM| = √(36+400+196)

|TM|= √(632)

|TM|= 25.139 = 25.14

4 0
3 years ago
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