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3 years ago

What is the nth term rule of the quadratic sequence below? 6,12,20,30,42,56,72

Mathematics

2 answers:

natali 33 [55]3 years ago

4 0

The ninth term is 110 beacsue you are adding two every time.

HOPE THIS HELPS!

HAVE A GR8 DAY ;-)

const2013 [10]3 years ago

3 0

Answer:

n(squared)^2 + 3n + 4.

Step-by-step explanation:

<h3>Sequence: 6, 12, 20, 30, 42, 56, 72.</h3>

First, lets find the first difference between the numbers. (In other words, let find the number gap difference between the first, second, third, fourth, fifth, sixth and seventh term).

In this case, it'll be..

first difference = 6.

second difference = 8.

third difference = 10.

fourth difference = 12.

fifth difference = 14.

sixth difference = 16.

Now we have the differences between the numbers, we need to do the second difference, which in this case is 3, meaning you have 3n.

Then, you would find out that n(squared)^2, is gonna be part one of your answer.

So, therefore so far we have in our answer: n^2 + 3n.

To then find the last part of our answer, we find our term 0. (The term before the first one given, in this case is 6).

So then you do you do your Term 1 = 6, and then take 2 away from it, giving you 4.

So, our final answer should be, n^2 + 3n + 4.

Hope :)

-Emilie Xo this is right and it helps! Xo

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Answer:

the probability that two 18 year old boys chosen at random will have heights greater than 185cm is 0.403

Step-by-step explanation:

P( x > 193) = 0.15

= 1- p(x less than or equal 193)

= 1 -p( z < (x- u) /sigma)

= 1- p( z< (193 - 187)/ sigma)

= 1- p( z< 6/ sigma)

P(z< 6/sigma) = 1 - 0.15

P(z < 6/sigma)= 0.85

6/sigma =1.036

Sigma= 6/1.036

Sigma= 5.79

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4 years ago

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lesya692 [45]

Answer:

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3 years ago

Uninhibited growth can be modeled by exponential functions other than A(t) =Upper A 0 e Superscript kt. For example, if an i

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:

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

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

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

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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