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IgorC [24]
2 years ago
14

What is the solution for r in the following system of equations? 2x + 3y = 16 2 + 2y = 10

Mathematics
1 answer:
klio [65]2 years ago
8 0

Answer:

(2,4)

*View attached graph*

Step-by-step explanation:

2x + 3y = 16

2 + 2y = 10

2 + 2y = 10

-2          - 2

2y = 8

/2  /2

y = 4

2x + 3y = 16

2x + 3(4) = 16

2x + 12 = 16

    - 12   - 12

2x = 4

/2   /2

x = 2

(x,y) -> (2,4)

Hope this helps!

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The Pacific halibut fishery has been modeled by the differential equation dy dt = ky 1 − y K where y(t) is the biomass (the tota
Dafna1 [17]

Answer:

a. The biomass weighs 2.30 * 10^7 kg after a year

b. It'll take 2.56 years to get to 4*10^7kg

Step-by-step explanation:

a.

k = 0.78,K = 6E7 kg

Given

dy/dt = ky(1- y/K)

Make ky dt the subject of formula

ky dt = dy/(1-y/K) --- make k dt the subject of formula

k dt = dy/(y(1-y/K))

k dt = K dy / y(K-y)

k dt = ((1/y) + (1/(K-y)))dy ---- integrate both sides

kt + c = ln(y/(K-y))

Ce^(kt) = y/(K-y)

Substitute the values of k and K

Ce^(0.78t) = y/(6*10^7 - y) ----- (1)

Given that y(0) = 2 * 10^7kg

(1) becomes

Ce^(0.78*0) = (2 * 10^7)/(6*10^7 - 2*10^7)

Ce° = (2*10^7)/(4*10^7

C = 2/7

Substitute 2/7 for C in (1)

2/7e^0.78t = y/(6*10^7 - y) ---(2)

We're to find the biomass a year later

So, t = 1

2/7e^0.78 = y/(6*10^7 - y)

0.62 = y/(6*10^7 - y)

y = 0.62(6*10^7 - y)

y = 0.62*6*10^7 - 0.62y

y + 0.62y = 0.62*6*10^7

1.62y = 0.62*6*10^7

1.62y = 3.72 * 10^7

y = 2.30 * 10^7kg.

Hence, the biomass weighs 2.30 * 10^7 kg after a year

b.

Here, we're to calculate the time it'll take the biomass to get to 4*10^7 kg

Substitute 4*10^7 for y in (2)

2/7e^0.78t = 4*10^7/(6*10^7 - 4*10^7)

2/7e^0.78t = 4*10^7/2*10^7

2/7e^0.78t = 2

e^0.78t = (2*7)/2

e^0.78t = 2

t = 2 * 1/0.78

t = 2.56 years

Hence, it'll take 2.56 years to get to 4*10^7kg

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