All categories

3 years ago

Solve for n. 2/3(1+n)=-1/2n

Mathematics

1 answer:

notka56 [123]3 years ago

5 0

Answer:

=−4/7

Step-by-step explanation:

And yh I'm sorry if I got it wrong

You might be interested in

Change each phrase into an algebraic expression

maks197457 [2]

Answer:

When translating phrases into algebraic expressions, you need to identify keywords and phrases which specifically refer to a mathematical operation (addition, subtraction, multiplication, and division). Usually, you can write out the algebraic expression of the verbal description in the order that it is said

3 0

2 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?

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:

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

3 0

3 years ago

Lily babysits on the weekends. She

Anastaziya [24]

Answer: She babysat 7 hours

Step-by-step explanation:if you take the knowledge you have; she charges 6$ per hour. and lily mad 42$ you can divide 42 by 6 to get the answer of 7

8 0

3 years ago

Read 2 more answers

Which algebraic expression represents this phrase? the product of 16 and the time after the start

8_murik_8 [283]

Answer:

product is 16 • t

Step-by-step explanation:

The first choice

3 0

3 years ago

Which are the two number that multiply is 300 and subtraction is 35

alukav5142 [94]

Answer:

15 and 20

Step-by-step explanation:

4 0

2 years ago