Ask question

Ask question

All categories

2 years ago

7

Insert parentheses to make the statement true 21-8-6=7

1 answer:

Advocard [28]2 years ago

5 0

(21-8)-6=7

and that's it

You might be interested in

Solve the right triangle. round your answer to the nearest tenth

mrs_skeptik [129]

Answer:

?

Step-by-step explanation:

4 0

3 years ago

Robin and evelyn are playing a target game. the object of the game is to get an object as close to the center as possible. each

svetlana [45]

Here's some of the info

~<span>Evelyn has the greater mean.

~</span>Evelyn’s objects generally landed farther away from the center than Robin’s.<span>
~</span>The object is to get the objects close to the center, so Robin is winning.<span>
</span>

5 0

3 years ago

Read 2 more answers

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

Help me with this problem show your work please

Akimi4 [234]

It cannot be simplified...

5 0

3 years ago

Sal Boxer decided to divide a gift of $5000 into two different accounts. He placed $1000 in an account that earns an annual simp

mash [69]

Answer:

THE TOTAL INTEREST EARNED = $ 385

Step-by-step explanation:

Interest earned by sal from first account can be calculated as follows :

Principal value (P) = $1000 .

interest rate (R) = 7.5 % .

Time period (T) = 1 year .

$\textbf{S . I} =\frac{\texbf{P * R * T}}{\textbf{100}}$

$= \frac{1000 * 7.5 * 1}{100}$

= $ 75 .

SIMILARLY , we can find the intrest earned by rest amount.

Principal value (P) = $5000 - $ 1000 = $ 4000 .

Interest rate (R) = 7.75 % .

Time period (T) = 1 year .

$S . I = \frac{4000 * 7.75 * 1}{100}$

= $ 310 .

THE TOTAL INTEREST EARNED = $ 75 + $ 310

= $ 385

4 0

3 years ago