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statuscvo [17]
3 years ago
12

Find the equation of the line that is perpendicular to the line y = (-1/3)x -1 and passes through the point (1, 5)?

Mathematics
1 answer:
Anit [1.1K]3 years ago
4 0

bearing in mind that perpendicular lines have negative reciprocal slopes, so


\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{10em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{1}{3}}x-1} \\\\[-0.35em] ~\dotfill


\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{1}\implies 3}}


so we're really looking for a line whose slope is 3 and runs through (1,5)


\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{5})~\hspace{10em} slope = m\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-5=3(x-1) \\\\\\ y-5=3x-3\implies y=3x+2

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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
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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

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