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Olegator [25]
3 years ago
9

Simplify. −5x4(−3x2+4x−2)

Mathematics
2 answers:
erastova [34]3 years ago
6 0

Answer:

15x^{6}-20x^{5}+10x^4)

Step-by-step explanation:

We have been given an expression -5x^4(-3x^2+4x-2) and we are asked to simplify our given expression.

We will use distributive property a(b+c)=a*b+a*c to simplify our given expression.

Upon distributing -5x^4 we will get,

(-5x^4*-3x^2)+(-5x^4*4x)-(-5x^4*2)

Using exponent property a^b*a^c=a^{b+c} we will get,

(15x^{4+2})+(-20x^{4+1})-(-10x^4)

(15x^{6})+(-20x^{5})-(-10x^4)

15x^{6}-20x^{5}+10x^4)

Therefore, our given expression simplifies to 15x^{6}-20x^{5}+10x^4).

Serggg [28]3 years ago
3 0
The answers my dear friends is 15x^6-20x^5+10x^4
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laila [671]

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?

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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} }

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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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