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3 years ago

The quotient of -15 and w

1 answer:

6 0

If you're dividing -15 by w then the answer would be -15/w. This is because the value of w has not been determined yet and a number cannot be divided by a variable.

If you're dividing w by -15, the same situation would occur due to the fact that a variable cannot be divided by a number. The answer would be w/-15.

A variable is merely a "placeholder", for lack of better terms, for a number(s) that have not been identified yet.

I hope I cleared this up!

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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

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

$r_A =3\%$ --- rate

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

Emma is training for a 10-kilometer race. She wants to beat her last 10-kilometer time, which was 1 hour 10 minutes. Emma has al

Whitepunk [10]

let x be how much longer she can run the race and still beat her previous time

55(min) + x(min) < 1(hr) + 10(min)

55(min) + x(min) < 60(min) + 10(min)

55 + x < 60 + 10

x < 60 + 10 - 55

x < 15(min)

7 0

3 years ago

TRUE OF FALSE

Keith_Richards [23]

Answer:

false it does not change the shape but it does change the size

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Please I really need help! I will mark you as brainliest!

iogann1982 [59]

Answer:

-10 / -5 / -1 / 0 / 1 / 5 / 10

Step-by-step explanation:

I hope this helps!

5 0

3 years ago

Determine whether the data described in the tables is best modeled by a linear function or an exponential function.

Leviafan [203]

Answer:

Electric car = Exponential function (decreasing cost)

Gasoline car = Exponential function (increasing cost)

Step-by-step explanation:

The given data the Electric car, we have;

Time (years) ${}$ Cost (thousands of dollars)

0 ${}$ 35

5 ${}$ 27

10 ${}$ 21

15 ${}$ 16

20 ${}$ 12.5

The difference between successive term is such that each term decreases by approximately the same percentage given as follows

1st and 2nd term (35 - 27)/35 × 100 ≈ 22.8571% decrease

2nd and 3rd term (27 - 21)/27 × 100 ≈ 22.222% decrease

3rd and 4th term (21 - 16)/21 × 100 ≈ 23.81% decrease

4th and 5th term (16 - 12.5)/16 × 100 ≈ 21.875% decrease

Therefore. the electric car is best approximated by exponential function

The given data the Gasoline car, we have;

Time (years) ${}$ Cost (thousands of dollars)

0 ${}$ 35

5 ${}$ 40.2

10 ${}$ 46.3

15 ${}$ 53.2

20 ${}$ 61.2

The difference between successive term is such that each term increases by approximately the same percentage given as follows

1st and 2nd term (40.2 - 35)/35 × 100 ≈ 14.857% decrease

2nd and 3rd term (46.3 - 40.2)/40.2 × 100 ≈ 15.174% decrease

3rd and 4th term (53.2 - 46.3)/46.3× 100 ≈ 14.9% decrease

4th and 5th term (61.2 - 53.2)/53.2× 100 ≈ 15.03% decrease

Therefore. the Gasoline car is best approximated by exponential function

4 0

3 years ago