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sveticcg [70]
2 years ago
13

What are the solutions of x^2-2x+5=0? A. X=2i or x=-2i. B. X=1+2i or x=1-2i. C. X=2+i or x=2-i. D. X=2+2i or x=2-2i

Mathematics
1 answer:
satela [25.4K]2 years ago
7 0

Answer:

B. x=1+2i or x=1-2i.

Step-by-step explanation:

The equation is given as;

x²-2x+5=0

complete the square method to find value of x as;

x²-2x = -5

x²-2x + 1 = -5 + 1

x²-2x +1 =-4

( x-1 )² = -4

x-1 = ±√-4 ---------<em>where √-1 = i</em>

x-1 =± 2i

So;

x-1 = + 2i

x= 1+2i

or

x= 1-2i

x= 1-2i

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The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than A(t)=A_{0}e^{kt}. for example, if an initial population P₀ requires n units of time to triple, then the function P(t)=P_{0}(3)^{\frac{t}{n} } models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form P(t)=P_{0}(3)^{\frac{t}{n} } that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form A(t)=A_{0}e^{kt}.

Answer: a) P(t)=50(3)^{\frac{t}{30} }

              b) P(t) = 280 insects

              c) t = 74 days

             d) A(t)=50e^{0.037t}

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

P(t)=50(3)^{\frac{t}{30} }

b) In t = 47 days:

P(t)=50(3)^{\frac{t}{30} }

P(47)=50(3)^{\frac{47}{30} }

P(47)=50(3)^{1.567}

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

750=50(3)^{\frac{t}{30} }

\frac{750}{50}=(3)^{\frac{t}{30} }

(3)^{\frac{t}{n} }=15

Using the property <u>Power</u> <u>Rule</u> of logarithm:

log(3)^{\frac{t}{30} }=log15

\frac{t}{30}log(3)=log15

t=\frac{log15}{log3} .30

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

A(t)=A_{0}e^{kt}

3A_{0}=A_{0}e^{30k}

3=e^{30k}

Use Power Rule again:

ln3=ln(e^{30k})

ln3=30k

k=\frac{ln3}{30}

k = 0.037

Equation for exponential growth will be:

A(t)=50e^{0.037t}

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