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fiasKO [112]
3 years ago
13

3y+11 plus 7x-30 plus 5x+14 find x and the y

Mathematics
1 answer:
Viktor [21]3 years ago
4 0
3y + 11 + 7x - 30 + 5x + 14

3y + 12x - 5

3y = -12x + 5

y = (-12x + 5)/3
x = (3y - 5)/-12

hope this helps
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Suppose a population grows according to the logistic equation but is subject to a constant total harvest rate of H. If N(t) is t
Elenna [48]

Answer:

a) Equilibrium point : [ 947, 53 ]

b) N = 947 is stable equilibrium, N = 53  is unstable equilibrium

c) N0, the population will not go extinct

Step-by-step explanation:

a)

Given that;

r = 2, k = 1000, H = 100

dN/dT = R(1 - N/k)N - H

so we substitute

dN/dt = 2( 1 - N/1000)N - 100

now for equilibrium solution, dN/dt = 0

so

2( 1 - N/1000)N - 100 = 0

((1000 - N)/1000)N = 50

N^2 - 1000N + 50000 = 0

N = 1000 ± √(-1000)² - 4(1)(50000)) / 2(1)

N = 947.213 OR 52.786

approximately

N = 947 OR 53

Therefore Equilibrium point : [ 947, 53 ]

b)

g(N) = 2( 1 - N/1000)N - 100

= 2N - N²/500 - 100

g'(N) = 2 - N/250

SO AT 947

g'(N) = g'(947) =  2 - 947/250 = -1.788 which is less than (<) 0

so N = 947 is stable equilibrium

now AT 53

g'(N) = g"(53) = 2 - 53/250 = 1.788 which is greater than (>) 0

so N = 53  is unstable equilibrium

The capacity k=1000

If the population is less than 53 then the population will become extinct but since the capacity is equal to 1000 then the population will not go extinct.

6 0
3 years ago
The ratio of the mass of an ant to the mass it can carry 1:50
slavikrds [6]

The ratio 1:50 means that an ant can carry an object heavy 50 times its mass.

So, if the mass of the ant is 0.4 grams, it can carry at most 50 times that mass, i.e. 50x0.4 = 20 grams

In order to answer the second question, you need to consider your mass and multiply it by 50: if you were as strong as an ant, you could lift that weight!

6 0
3 years ago
What is the answer please help no links thank you
Nana76 [90]

Answer:

x=1

Step-by-step explanation:

brainliest please

7 0
3 years ago
Use the intermediate value theorem to find the value of c such that f(c) = M. f(x) = x^2 - x + 1 text( on ) [1,8]; M = 21 c =
DanielleElmas [232]

Answer:

c = 5

Step-by-step explanation:

Given

f(c) = M

f(x) = x^2 - x + 1

Interval: [1,8]

M = 21

Required

Find c using Intermediate Value theorem

First, check if the value of M is within the given range:

f(x) = x^2 - x + 1

f(1) = 1^2 - 1 + 1

f(1) = 1

f(x) = 8^2 - 8 + 1

f(x) = 57

1 \le M \le 57

1 \le 21 \le 57

M is within range.

Solving further:

We have:

f(c) = f(x) = M

f(x) = 21

Substitute 21 for f(x) in f(x) = x^2 - x + 1

21 = x^2 - x + 1

Express as quadratic function

x^2 - x + 1 - 21  = 0

x^2 - x - 20  = 0

Expand

x^2 + 4x - 5x - 20

x(x+4)-5(x+4)=0

(x - 5)(x+4) = 0

x - 5 = 0 or x + 4= 0

x = 5 or x = -4

The value of x = -4 is outside the Interval: [1,8]

So:

x = 5

f(c) = f(x) = M

f(c) = f(5) = 21

By comparison:

c = 5

4 0
3 years ago
What is 2.63 repeating as a mixed number in simplest form
galben [10]
2 63/100

(2 and 63 hundredths)
6 0
3 years ago
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