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Ganezh [65]
3 years ago
7

Can somebody help me with the problem in the picture :( pleasee

Mathematics
2 answers:
Alisiya [41]3 years ago
7 0

Answer:

2^3*3*5=120

Step-by-step explanation:

5 is the prime number

monitta3 years ago
5 0
I think the answer is 5
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How do I solve this problem −24−18p=38p?
Leni [432]
P would equal -3/7
Hope it helps!

5 0
4 years ago
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135 + 230+100+100+1000​
PIT_PIT [208]

Answer:

its 1565!

Step-by-step explanation:

8 0
3 years ago
Identify the contrapositive of the following statement. If x = 2, then x + 3 = 5.
antiseptic1488 [7]

Answer:

If x + 3 ≠ 5, then x ≠ 2 thus b: is your Answer

Step-by-step explanation:

4 0
4 years ago
Which numbers make the inequality true? Select all that apply. 7 < _____ < 8
jekas [21]

Answer:

7<√56<8

(7)^2<(√56)^2<(8)^2

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6 0
2 years ago
Consider the following predator-prey systems of differential equations dx/dt = -x/2 + xy/2 + y, dy/dt = y(1 - y) - xy/2 + y.
fgiga [73]

Answer:

a) Prey: \frac{dy}{dt}=y(1-y)-\frac{xy}{2+y}

Predator: \frac{dx}{dt}=\frac{xy}{2+y}-\frac{x}{2}

b) they grow as a function y(1-y).

c) It will grow faster and probably exponentially,

Step-by-step explanation:

a)

Prey

For prey population the rate of change is given by its own growth rate, it would be a constant times the variable, minus the rate at which it is preyed upon, usually it is represent a parameter times x and y.

So the differential equation could be:

\frac{dy}{dt}=y(1-y)-\frac{xy}{2+y}

Predator

By the other hand, the rate of change of the predator's population depends only the rate at which the predators eat preys, minus predator's intrinsic death rate.

\frac{dx}{dt}=\frac{xy}{2+y}-\frac{x}{2}

b)  If there are no predators present the differential equation that models prey population will be:

\frac{dy}{dt}=y(1-y)

So they grow as a function y(1-y).

c)  In the case of the population of prey is much grader than predator, it will grow faster and exponentially, because it we assume that preys have unlimited food supply.

I hope it helps you!

7 0
4 years ago
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