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Lina20 [59]
3 years ago
6

(x^2y^3) = (xy^a)^b

Mathematics
1 answer:
Rainbow [258]3 years ago
4 0

Answer:

C. \frac{3}{2}

Step-by-step explanation:

To find the value f b, we need to compare the exponents.

The given exponential equation is:

( {x}^{2}  {y}^{3} )^{3}  = ( {x}  {y}^{a} )^{b}

Recall and apply the following rule of exponents.

( {x}^{m}  )^{n}  = {x}^{mn}

We apply this rule on both sides to get:

{x}^{2 \times 3}  {y}^{3 \times 3} =   {x}^{b}  {y}^{ab}

Simplify the exponents on the left.

{x}^{6}  {y}^{9} =   {x}^{b}  {y}^{ab}

Comparing exponents of the same variables on both sides,

b = 6 \: and \:\: ab = 9

\implies \: 6b = 9

Divide both sides by 6.

b =  \frac{9}{6}

b =  \frac{3}{2}

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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
Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin
ozzi

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by using the integration formula

we get,

\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c

now put the value of t=\sin\theta in the above equation

we get,

\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c

hence proved

7 0
3 years ago
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