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loris [4]
3 years ago
8

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

Mathematics
1 answer:
tangare [24]3 years ago
5 0

Answer:

-1 -3

Step-by-step explanation:

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Suppose you receive an allowance of $107 and your friend receives an allowance of $83 at the beginning of the month. If you spen
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Y= 107-8x
Y=83-4x
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Over to the left
83+ 4x=107

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3 years ago
On a piece of paper, draw a box plot to represent the data below. Then
777dan777 [17]
Since they are already in order you know the 22 and the 61 are the upper and lower extremes. To find the median you find the number that is in the middle which is 42. To find the lower quartile find the middle number starting from the first number and the number to the left of the median (22 and 36 in this case). The loser quartile is 25. To find the upper quartile, fine the middle number between the number to the right of the median and the last number (44 and 61 in this case.) the upper quartile is 57 you then just plot it and graph it

The work is attached

6 0
3 years ago
Hello - I won't be able to go to bed if I don't solve this thing.
Darina [25.2K]
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7 0
3 years ago
Read 2 more answers
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
Please I truly need help I’m not that good with math
m_a_m_a [10]

Answer:

the answer is A

Step-by-step explanation:

np

4 0
3 years ago
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