The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 42 and the common ratio is three fourths. Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.

Answers

Answer 1
Answer:

Answer:

The sum in sigma form is \sum_(i=1)^(\infty)42((3)/(4))^(i)

The upper limit of the population is 168.

Step-by-step explanation:

We are given that,

Population of dragonfly is represented by the series with,

First term, a_(1)=42

Common ratio, r=(3)/(4)

So, we see that,

The sum in sigma form is given by \sum_(i=1)^(\infty)a_(1)r^(i)

That is, \sum_(i=1)^(\infty)42((3)/(4))^(i)

Now, the infinite sum of the series is S=(a_1)/(1-r)

So, the sum is S=(42)/(1-(3)/(4))

i.e. S=(42* 4)/(4-3)

i.e. S=(168)/(1)

Thus, the upper limit of the population is 168.

Answer 2
Answer: Hello,

\lim_(n \to \infty) \sum_(i=0)^(i=n) a_1*((3)/(4))^i =42*4=168

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What is the solution to the equation 3/4 (4c + 16)= 2c+9

c=

Answers

If you would like to solve the equation 3/4 * (4 * c + 16) = 2 * c + 9, you can calculate this using the following steps:

3/4 * (4 * c + 16) = 2 * c + 9
3/4 * 4 * c + 3/4 * 16 = 2 * c + 9
3 * c + 12 = 2 * c + 9
3 * c - 2 * c = 9 - 12
c = - 3

The correct result would be - 3.
1st -- multiplying the left side = 3c+12 which equals to 2c+9 then subtracting 2c from both sides which equal to the c=-3. Answer is -3.

Alon and his friend Marna own a chimney sweep service company. Working together, they can clean a 20-foot chimney in 1 Five-sevenths hours. If it takes Kalon 4 hours to clean a 20-foot chimney by himself, how long does it take Marna to clean the same size chimney by herself?

Answers

Answer:

3 hours

Step-by-step explanation:

Given that:

Time taken by Kalon = 4 hours

Time taken by Marna =?

Time taken if the work together = 1 5/7 hours

Kalon's rate = 1/ time =. 1/4

Kalon + Marna = 1 ÷ 1 5/7 = 7/12

Hence,

Marna's rate = 7/12 - 1/4

Marna's Rate = (7 - 3) / 12 = 4/12 = 1/3

Marna's rate = 1/3

Hence, time it takes Marna = reciprocal of rate = 1÷ 1/3 = 1 * 3/1 = 3 hours

Answer:

3 hours

Step-by-step explanation:

Solve the equation. 3/2 + b = 7/4 (stylised in fractions)

b= ?

Answers

3/2 + b = 7/4

First rewrite 3/2 to have the same denominator as 7/4:

3/2 x 2 = 6/4

Now you have 6/4 + b = 7/4

To find b subtract 6/4 from 7/4:

b = 7/4 - 6/4

b = 1/4

An entertainment firm offers several DJ choices and light shows that range in price based on the rental time period. The DJ's cost between $219.00 and $369.00 per night and the light shows cost between $159.00 and $309.00 per night. If you are booking both a DJ and a light show, write a compound inequality that represents the possible total amount you would pay, x.

Answers

219.00 + 159.00 = 378.00, the lowest cost of DJ & lights 369.00 + 309.00 = 678.00, the highest cost of DJ & lights This means the lowest cost would be $434.00, and the highest cost would be $678.00, which written as our compound inequality: $378.00 ≤ x ≤ $678.00 If we wanted to write it as an interval format, it would look like: [$378.00, $678.00] (We use brackets because it's less/greater than or equal to those amounts.)

Use the Change of Base Formula to evaluate log3 58. Then convert log3 58 to a logarithm in base 4. Round to the nearest thousandth.

Answers

Answer:

\log_358\approx 3.696

\Rightarrow (\log_458)/(\log_43)

Step-by-step explanation:

Given: \log_358

We need to re-write the log expression with base 4 using base change property of log.

Log property:

\log_ab=(\log_cb)/(\log_ca)

Evaluate:

\Rightarrow \log_358

\Rightarrow (\log58)/(\log3)

\Rightarrow (1.763427)/(0.47712)\approx 3.696

Convert with base 4:

\Rightarrow \log_358

\Rightarrow (\log_458)/(\log_43)

Hello,

log_3(58)= (ln(58))/(ln(3))=3,69597450... \n log_3(58)=log_4(x)\n (ln(58))/(ln(3)) = (ln(x))/(4)==\textgreater\ x=e^{ (ln(58)*ln(4))/(ln(3))}\n =167,957104437222249066099...

≈167.96

Which number is a common factor of 24, 48, and 144?16

10

9

8

Answers

8 is the common factor 24÷8=3 48÷8=6 144÷8=18