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3 years ago

RESTAURANTS In 2012, a popular pizza franchise had 2483 restaurants. In 2017, there were 2606 franchised

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

5 0

y - 2483 = 24.6(x - 2)

Let x represent the number of years after 2010

Let y represents the number of restaurants in x years

2012 is 2 years after 2010

2017 is 7 years after 2010

Number of restaurants owned by RESTAURANTS in 2012 = 2483

This can be represented in coordinate form as (2, 2483)

Number of restaurants owned by RESTAURANTS in 2017 = 2606

This can be represented in coordinate form as (7, 2606)

The equation of a line in point-slope form is:

$y-y_{1} = m(x - x_{1} )$

where m represents this slope, and is calculated as:

$m = \frac{y_{2} - y_{1} }{x_2-x_1} \\m = \frac{2606-2483}{7-2} \\m = \frac{123}{5} \\m = 24.6$

Substitute x₁ = 2, y₁ = 2483, and m = 24.6 into the point-slope equation

The equation in point-slope form becomes:

y - 2483 = 24.6(x - 2)

Learn more here: brainly.com/question/6497976

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It takes Natalie five hours to tar a roof. Imani can tar the same roof in ten hours. How long would it take them if they worked

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Answer:

C. 3.33 hours

Step-by-step explanation:

Use the algebra work equation:

$\frac{1}{tb}$ = $\frac{1}{t1}$ + $\frac{1}{t2}$ , where tb is the time to work together, t1 is the time it takes one person, and t2 is the time it takes the other person

Plug in the values we know:

$\frac{1}{tb}$ = $\frac{1}{5}$ + $\frac{1}{10}$

$\frac{1}{tb}$ = $\frac{4}{20}$ + $\frac{2}{20}$

$\frac{1}{tb}$ = $\frac{6}{20}$

20 = 6tb

3.33 = tb

So, it would take them 3.33 hours when working together.

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3 years ago

Persons A and B are at the beach, their eyes are 5 ft and 6 ft, respectively, above sea level. How many miles farther out is Per

Marysya12 [62]

Answer:

The radius of the earth is something like 4000 miles= about 20,000,000 feet. (Look up the exact radius of the earth to calculate the correct number yourself, I'm too lazy!)

The horizon is where you're looking at a tangent to the earth's surface. Since the tangent to a circle is perpendicular to the radius (if we assume the earth is a perfect sphere), we can set up a right angled triangle:

H...P

Where C is the centre of the earth, H is the horizon, and P is the person. CH = 20,000,000 feet. CP = 20,000,005 feet (in the case of the 5' person, that'd be me :) And there is a right angle at H.

So using Pythagoras' theorem, if d = PH, then:

d^2 = 20,000,005^2 - 20,000,000^2

=> d^2 = 200,000,025

=> d = 14142 feet (just under 3 miles)

For a 6' person:

d^2 = 20,000,006^2 - 20,000,000^2

=> d^2 = 240,000,036

=> d = 15491 feet

So the difference is 1349 feet or 0.256 miles

Repeat this exercise with the real radius of the earth (in feet) to get the correct answer, but it'll be pretty close to what I've given here

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3 years ago

Evaluate each expression if w = 2, x = 3, y = 5, and z = 6.<br><br><br> z/2=

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Step-by-step explanation:

We plug in 6 for z

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3 years ago

Read 2 more answers

A yo-yo is moving up and down a string so that its velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. The initial pos

jeka57 [31]

Part A - The average value of v(t) over the interval (0, π/2) is 6/π

Part B - The displacement of the yo-yo from time t = 0 to time t = π is 0 m

Part C - The total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

<h3>Part A: Find the average value of v(t) on the interval (0, π/2)</h3>

The average value of a function f(t) over the interval (a,b) is

$f(t)_{avg} = \frac{1}{b - a} \int\limits^b_a {f(t)} \, dx$

So, since velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. Its average value over the interval (0, π/2) is given by

$v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {v(t)} \, dt$

Since v(t) = 3cost, we have

$v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {3cos(t)} \, dt\\= \frac{3}{\frac{\pi }{2}} \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= \frac{6}{{\pi}} [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= \frac{6}{{\pi}} [{sin(\frac{\pi }{2})} - sin0]\\ = \frac{6}{{\pi}} [1 - 0]\\ = \frac{6}{{\pi}} [1]\\ = \frac{6}{{\pi}}$

So, the average value of v(t) over the interval (0, π/2) is 6/π

<h3>Part B: What is the displacement of the yo-yo from time t = 0 to time t = π?</h3>

To find the displacement of the yo-yo, we need to find its position.

So, its position x = ∫v(t)dt

= ∫3cos(t)dt

= 3∫cos(t)dt

= 3sint + C

Given that at t = 0, x = 3. so

x = 3sint + C

3 = 3sin0 + C

3 = 0 + C

C = 3

So, x(t) = 3sint + 3

So, its displacement from time t = 0 to time t = π is

Δx = x(π) - x(0)

= 3sinπ + 3 - (3sin0 + 3)

= 3 × 0 + 3 - 0 - 3

= 0 + 3 - 3

= 0 + 0

= 0 m

So, the displacement of the yo-yo from time t = 0 to time t = π is 0 m

<h3>Part C: Find the total distance the yo-yo travels from time t = 0 to time t = π. (10 points)</h3>

The total distance the yo-yo travels from time t = 0 to time t = π is given by

$x(t) = \int\limits^{\pi}_0 {v(t)} \, dt\\= \int\limits^{\pi }_0 {3cos(t)} \, dt\\= 3 \int\limits^{\pi }_0 {cos(t)} \, dt\\ = 3 \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt + 3\int\limits^{\pi }_{\frac{\pi }{2}} {cos(t)} \, dt\\= 3 \times 2\int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= 6 [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= 6[{sin\frac{\pi }{2} - sin0]\\\\= 6[1 - 0]\\= 6(1)\\= 6$

So, the total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

Learn more about average value of a function here:

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1 year ago

If f(x)= 3x + 1 and f^(-1)= x-1/3, then the ordered pair of f^(-1)(10)=

iragen [17]

The answer is B

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plug in 10 to the inverse

10/3 - 1/3 = 9/3 or 3

this gives you the point (10,3)

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3 years ago