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3 years ago

6

2-^3= -8 -6 -1/6 1/8 2/3

1 answer:

Allisa [31]3 years ago

8 0

Answer:

-8

Step-by-step explanation:

Hope this helps!

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City A had a population of 10000 in the year 1990. City A’s population grows at a constant rate of 3% per year. City B has a pop

Georgia [21]

Answer:

City A and city B will have equal population 25years after 1990

Step-by-step explanation:

Given

Let

$t \to$ years after 1990

$A_t \to$ population function of city A

$B_t \to$ population function of city B

<u>City A</u>

$A_0 = 10000$ ---- initial population (1990)

$r_A =3\%$ --- rate

<u>City B</u>

$B_{10} = \frac{1}{2} * A_{10}$ ----- t = 10 in 2000

$A_{20} = B_{20} * (1 + 20\%)$ ---- t = 20 in 2010

Required

When they will have the same population

Both functions follow exponential function.

So, we have:

$A_t = A_0 * (1 + r_A)^t$

$B_t = B_0 * (1 + r_B)^t$

Calculate the population of city A in 2000 (t = 10)

$A_t = A_0 * (1 + r_A)^t$

$A_{10} = 10000 * (1 + 3\%)^{10}$

$A_{10} = 10000 * (1 + 0.03)^{10}$

$A_{10} = 10000 * (1.03)^{10}$

$A_{10} = 13439.16$

Calculate the population of city A in 2010 (t = 20)

$A_t = A_0 * (1 + r_A)^t$

$A_{20} = 10000 * (1 + 3\%)^{20}$

$A_{20} = 10000 * (1 + 0.03)^{20}$

$A_{20} = 10000 * (1.03)^{20}$

$A_{20} = 18061.11$

From the question, we have:

$B_{10} = \frac{1}{2} * A_{10}$ and $A_{20} = B_{20} * (1 + 20\%)$

$B_{10} = \frac{1}{2} * A_{10}$

$B_{10} = \frac{1}{2} * 13439.16$

$B_{10} = 6719.58$

$A_{20} = B_{20} * (1 + 20\%)$

$18061.11 = B_{20} * (1 + 20\%)$

$18061.11 = B_{20} * (1 + 0.20)$

$18061.11 = B_{20} * (1.20)$

Solve for B20

$B_{20} = \frac{18061.11}{1.20}$

$B_{20} = 15050.93$

$B_{10} = 6719.58$ and $B_{20} = 15050.93$ can be used to determine the function of city B

$B_t = B_0 * (1 + r_B)^t$

For: $B_{10} = 6719.58$

We have:

$B_{10} = B_0 * (1 + r_B)^{10}$

$B_0 * (1 + r_B)^{10} = 6719.58$

For: $B_{20} = 15050.93$

We have:

$B_{20} = B_0 * (1 + r_B)^{20}$

$B_0 * (1 + r_B)^{20} = 15050.93$

Divide $B_0 * (1 + r_B)^{20} = 15050.93$ by $B_0 * (1 + r_B)^{10} = 6719.58$

$\frac{B_0 * (1 + r_B)^{20}}{B_0 * (1 + r_B)^{10}} = \frac{15050.93}{6719.58}$

$\frac{(1 + r_B)^{20}}{(1 + r_B)^{10}} = 2.2399$

Apply law of indices

$(1 + r_B)^{20-10} = 2.2399$

$(1 + r_B)^{10} = 2.2399$ --- (1)

Take 10th root of both sides

$1 + r_B = \sqrt[10]{2.2399}$

$1 + r_B = 1.08$

Subtract 1 from both sides

$r_B = 0.08$

To calculate $B_0$ , we have:

$B_0 * (1 + r_B)^{10} = 6719.58$

Recall that: $(1 + r_B)^{10} = 2.2399$

So:

$B_0 * 2.2399 = 6719.58$

$B_0 = \frac{6719.58}{2.2399}$

$B_0 = 3000$

Hence:

$B_t = B_0 * (1 + r_B)^t$

$B_t = 3000 * (1 + 0.08)^t$

$B_t = 3000 * (1.08)^t$

The question requires that we solve for t when:

$A_t = B_t$

Where:

$A_t = A_0 * (1 + r_A)^t$

$A_t = 10000 * (1 + 3\%)^t$

$A_t = 10000 * (1 + 0.03)^t$

$A_t = 10000 * (1.03)^t$

and

$B_t = 3000 * (1.08)^t$

$A_t = B_t$ becomes

$10000 * (1.03)^t = 3000 * (1.08)^t$

Divide both sides by 10000

$(1.03)^t = 0.3 * (1.08)^t$

Divide both sides by $(1.08)^t$

$(\frac{1.03}{1.08})^t = 0.3$

$(0.9537)^t = 0.3$

Take natural logarithm of both sides

$\ln(0.9537)^t = \ln(0.3)$

Rewrite as:

$t\cdot\ln(0.9537) = \ln(0.3)$

Solve for t

$t = \frac{\ln(0.3)}{ln(0.9537)}$

$t = 25.397$

Approximate

$t = 25$

7 0

3 years ago

Parisian sewer #1 contains 9 fewer gigantic rats—rats as big as cats, some might say—than parisian sewer #2. parisian sewer #2 c

Leya [2.2K]

There are different kinds of math problem. There will be 11 rats in sewer #1.

<h3>What are word problem?</h3>

The term word problems is known to be problems that are associated with a story, math, etc. They are known to often vary in terms of technicality.

Lets take

sewer #1 = a

sewer #2 = b

sewer #3 = c

Note that A=B-9

So then you would have:

A=B-9

B=C- 5

A+B+C=56

Then you have to do a substitution so as to find C:

(B- 9) + (C-5) + C = 56

{ (C- 5)-9} + (C-5) + C = 56

3C - 19 = 56

3C = 75

B = C- 5

B = 25 - 5

Therefore, B = 20

A = B - 9

= 25 - 9

=11

Therefore, there are are 11 rats in sewer #1

Learn more about Word Problems from

brainly.com/question/21405634

8 0

2 years ago

In a classroom of 20 students, 70% of them have brown eyes. If a teacher picks two students at random to pass out supplies, what

antiseptic1488 [7]

Answer:

6/20 can I get brainliest pleast

Step-by-step explanation:

$\frac{70}{100} \times 20 = 14 \\ 20 - 14 = 6 \\$

there is a chance that the teacher will pick a student without brown eyes 6/20 times

4 0

3 years ago

3 1/2+ 5 3/8 can you solve please

Sunny_sXe [5.5K]

Answer:

71/8

8.875

8 7/8

Step-by-step explanation:

8 0

4 years ago

The difference between thirty and five

Rudiy27

The difference bettween 30 and 5 would be 25

5 0

3 years ago

Read 2 more answers