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Jobisdone [24]
1 year ago
9

I need this equation solved

Mathematics
1 answer:
goldfiish [28.3K]1 year ago
6 0

Answer:

28

Step-by-step explanation:

similar triangles are proportional

and add all the sides of the bigger triangle

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Combine these radicals<br> 3V2-5V2<br><br> Plz help<br> Will give Brainlyest
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Answer:

b

Step-by-step explanation:

because i said so and because i am tired and tired and tired

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Marie has a part time job. She earns $7 an hour. She makes at most $143.50 per week. What is the greatest number of hours that s
Troyanec [42]
We can divide 143.50 by 7. This gives us 20.5. This means that she works 20.5 hours a week or 21 hours if you round up.
7 0
3 years ago
Angles BAE and FAC are straight angles. What angle relationship best describes angles BAC and EAF?
Vika [28.1K]

The best angle relationship that describes angles BAC and EAF is supplementary angles

The sum of angle on a straight line is supplementary i.e. they sum up to 180 degrees.

If Angles BAE and FAC are straight angles, it means they are linear pairs and their sum is 180 degrees. Mathematically;

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7 0
3 years ago
3 museums charge an entrance fee based on the number of visitors in the group. the tables list the fees charged by the museums.
miv72 [106K]
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3 years ago
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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

8 0
3 years ago
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