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IRISSAK [1]
3 years ago
12

How would the fraction

Mathematics
2 answers:
Bumek [7]3 years ago
8 0

Answer:

A. {-\frac{5+5\sqrt 3 }{2}}

Step-by-step explanation:

Please wait for more explanation:

\frac{5}{1-\sqrt 3}=  \frac{5}{1-\sqrt 3}\times  \frac{1+\sqrt 3}{1+\sqrt 3}\\\\= \frac{5(1+\sqrt 3) }{1^2 -(\sqrt 3) ^2}\\\\= \frac{5+5\sqrt 3 }{1 -3}\\\\= \frac{5+5\sqrt 3 }{-2}\\\\=\huge\purple {\boxed {-\frac{5+5\sqrt 3 }{2}}} \\

aleksley [76]3 years ago
5 0

The differences of squares:  a² - b² = (a - b)(a + b)

at denominator we have:  a - b  and we want:  a² - b²

so we need to multiply by (a+b)  {both denominator and nominator!}

\dfrac5{1-\sqrt3}=\dfrac{5(1+\sqrt3)}{(1-\sqrt3)(1+\sqrt3)}=\dfrac{5(1+\sqrt3)}{1^2-(\sqrt3)^2}=\dfrac{5+5\sqrt3}{1-3}=\dfrac{5+5\sqrt3}{-2}=-\dfrac{5+5\sqrt3}{2}

The answer is  A

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Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and
zloy xaker [14]

Answer:

(0,0)   (4000,0) and (500,79)

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW

\frac{dW}{dt} = -0.02W + 0.00004RW

To solve this, we first equate \frac{dR}{dt} and \frac{dW}{dt} to 0.

So, we have:

0.09R(1 - 0.00025R) - 0.001RW = 0

-0.02W + 0.00004RW = 0

Factor out R in 0.09R(1 - 0.00025R) - 0.001RW = 0

R(0.09(1 - 0.00025R) - 0.001W) = 0

Split

R = 0   or 0.09(1 - 0.00025R) - 0.001W = 0

R = 0   or  0.09 - 2.25 * 10^{-5}R - 0.001W = 0

Factor out W in -0.02W + 0.00004RW = 0

W(-0.02 + 0.00004R) = 0

Split

W = 0 or -0.02 + 0.00004R = 0

Solve for R

-0.02 + 0.00004R = 0

0.00004R = 0.02

Make R the subject

R = \frac{0.02}{0.00004}

R = 500

When R = 500, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0

0.09 -0.01125 - 0.001W = 0

0.07875 - 0.001W = 0

Collect like terms

- 0.001W = -0.07875

Solve for W

W = \frac{-0.07875}{ - 0.001}

W = 78.75

W \approx 79

(R,W) \to (500,79)

When W = 0, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0

0.09 - 2.25 * 10^{-5}R = 0

Collect like terms

- 2.25 * 10^{-5}R = -0.09

Solve for R

R = \frac{-0.09}{- 2.25 * 10^{-5}}

R = 4000

So, we have:

(R,W) \to (4000,0)

When R =0, we have:

-0.02W + 0.00004RW = 0

-0.02W + 0.00004W*0 = 0

-0.02W + 0 = 0

-0.02W = 0

W=0

So, we have:

(R,W) \to (0,0)

Hence, the points of equilibrium are:

(0,0)   (4000,0) and (500,79)

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Answer:

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Step-by-step explanation:

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5 0
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