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jek_recluse [69]
3 years ago
7

The quotient of -15 and w

Mathematics
1 answer:
yaroslaw [1]3 years ago
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!
You might be interested in
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
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
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