Question

part 1: matching scenarios, tables, and equations
directions: cut out the following cards and match one scenario to one table, equation, and solution. fill in the missing values in the tables. show your work on the card or on a separate piece of paper.

a
a certain ceiling is made up of tiles. every square meter of ceiling requires 10.75 tiles. how many tiles are needed for a ceiling that is 28.5 square meters?

d
jenny is painting her house. after 6 hours, she has used 2 gallons of paint. how many hours will it take jenny to use 5 gallons of paint?

g

square meters	number of tiles
	7
6	10.5
x

b
$y = \frac{3}{4}x$

e

square meters	number of tiles
8	86
	129
x

h
$y = 3x$

c
solution:
80

f
$y = 1.75x$

i
$y = 4x$

Explanation:

To solve the problem of finding the number of tiles for a 28.5 - square - meter ceiling (Scenario A), we can use the following steps:

Step 1: Identify the relationship

We know that for every square meter of the ceiling, 10.75 tiles are required. This means that the number of tiles \(y\) is directly proportional to the area of the ceiling in square meters \(x\), and the relationship can be expressed as \(y = 10.75x\)

Step 2: Substitute the given value of \(x\)

We are given that the area of the ceiling \(x=28.5\) square meters. We substitute \(x = 28.5\) into the equation \(y=10.75x\)

$$y=10.75\times28.5$$

Step 3: Calculate the product

First, we multiply \(10.75\) and \(28.5\):

$$10.75\times28.5=(10 + 0.75)\times28.5=10\times28.5+0.75\times28.5$$

$$=285+21.375$$

$$=306.375$$

So the number of tiles required for a 28.5 - square - meter ceiling is \(306.375\)

For the table in card E (relating square meters to number of tiles with the equation \(y = 10.75x\)):

• When \(x = 1\), \(y=10.75\times1 = 10.75\)
• When \(y = 129\), we solve for \(x\) from \(y = 10.75x\), so \(x=\frac{y}{10.75}=\frac{129}{10.75} = 12\)
• For a general \(x\), \(y = 10.75x\)

For the table in card G (let's assume the equation is \(y = 1.75x\) as \(y = 1.75x\) when \(x = 6\), \(y=1.75\times6 = 10.5\) which matches the table):

• When \(x = 1\), \(y=1.75\times1=1.75\)
• When \(y = 7\), we solve for \(x\) from \(y = 1.75x\), so \(x=\frac{y}{1.75}=\frac{7}{1.75}=4\)
• For a general \(x\), \(y = 1.75x\)

For Scenario D (Jenny painting her house):

• We know that the rate of using paint is \(\frac{2\space gallons}{6\space hours}=\frac{1}{3}\space gallons\space per\space hour\)
• Let \(t\) be the time in hours and \(g\) be the number of gallons. The relationship is \(g=\frac{1}{3}t\) or \(t = 3g\)
• When \(g = 5\) gallons, \(t=3\times5 = 15\) hours

If we consider the equation - table - scenario matching:

• Scenario A (ceiling tiles) matches with the equation that has a constant of proportionality of \(10.75\) (not one of the given simple equations like \(y=\frac{3}{4}x,y = 3x,y=1.75x,y = 4x\) directly, but the calculation for the number of tiles is as above)
• Scenario D (Jenny painting) has a rate of \(\frac{2\space gallons}{6\space hours}=\frac{1}{3}\space gallons\space per\space hour\), and if we rewrite the time - gallon relationship, \(t = 3g\), which matches the equation \(y = 3x\) (where \(x\) is gallons and \(y\) is time)
• The table in G with the equation \(y = 1.75x\) (since \(1.75\times6=10.5\) which matches the table)
• The table in E with the equation \(y = 10.75x\) (since \(10.75\times8 = 86\) and \(10.75\times12=129\))

If we are to fill the missing values:

For Table G (\(y = 1.75x\)):

• When \(x = 1\), \(y=1.75\times1 = 1.75\)
• When \(y = 7\), \(x=\frac{7}{1.75}=4\)
• For a general \(x\), \(y = 1.75x\)

For Table E (\(y = 10.75x\)):

• When \(x = 1\), \(y = 10.75\)
• When \(y = 129\), \(x=\frac{129}{10.75}=12\)
• For a general \(x\), \(y = 10.75x\)

For Scenario D:

If \(g = 5\) gallons, \(t=3\times5 = 15\) hours

For Scenario A:

The number of tiles \(y = 10.75\times28.5=306.375\)

Answer:

To solve the problem of finding the number of tiles for a 28.5 - square - meter ceiling (Scenario A), we can use the following steps:

Step 1: Identify the relationship

We know that for every square meter of the ceiling, 10.75 tiles are required. This means that the number of tiles \(y\) is directly proportional to the area of the ceiling in square meters \(x\), and the relationship can be expressed as \(y = 10.75x\)

Step 2: Substitute the given value of \(x\)

We are given that the area of the ceiling \(x=28.5\) square meters. We substitute \(x = 28.5\) into the equation \(y=10.75x\)

$$y=10.75\times28.5$$

Step 3: Calculate the product

First, we multiply \(10.75\) and \(28.5\):

$$10.75\times28.5=(10 + 0.75)\times28.5=10\times28.5+0.75\times28.5$$

$$=285+21.375$$

$$=306.375$$

So the number of tiles required for a 28.5 - square - meter ceiling is \(306.375\)

For the table in card E (relating square meters to number of tiles with the equation \(y = 10.75x\)):

• When \(x = 1\), \(y=10.75\times1 = 10.75\)
• When \(y = 129\), we solve for \(x\) from \(y = 10.75x\), so \(x=\frac{y}{10.75}=\frac{129}{10.75} = 12\)
• For a general \(x\), \(y = 10.75x\)

For the table in card G (let's assume the equation is \(y = 1.75x\) as \(y = 1.75x\) when \(x = 6\), \(y=1.75\times6 = 10.5\) which matches the table):

• When \(x = 1\), \(y=1.75\times1=1.75\)
• When \(y = 7\), we solve for \(x\) from \(y = 1.75x\), so \(x=\frac{y}{1.75}=\frac{7}{1.75}=4\)
• For a general \(x\), \(y = 1.75x\)

For Scenario D (Jenny painting her house):

• We know that the rate of using paint is \(\frac{2\space gallons}{6\space hours}=\frac{1}{3}\space gallons\space per\space hour\)
• Let \(t\) be the time in hours and \(g\) be the number of gallons. The relationship is \(g=\frac{1}{3}t\) or \(t = 3g\)
• When \(g = 5\) gallons, \(t=3\times5 = 15\) hours

If we consider the equation - table - scenario matching:

• Scenario A (ceiling tiles) matches with the equation that has a constant of proportionality of \(10.75\) (not one of the given simple equations like \(y=\frac{3}{4}x,y = 3x,y=1.75x,y = 4x\) directly, but the calculation for the number of tiles is as above)
• Scenario D (Jenny painting) has a rate of \(\frac{2\space gallons}{6\space hours}=\frac{1}{3}\space gallons\space per\space hour\), and if we rewrite the time - gallon relationship, \(t = 3g\), which matches the equation \(y = 3x\) (where \(x\) is gallons and \(y\) is time)
• The table in G with the equation \(y = 1.75x\) (since \(1.75\times6=10.5\) which matches the table)
• The table in E with the equation \(y = 10.75x\) (since \(10.75\times8 = 86\) and \(10.75\times12=129\))

If we are to fill the missing values:

For Table G (\(y = 1.75x\)):

• When \(x = 1\), \(y=1.75\times1 = 1.75\)
• When \(y = 7\), \(x=\frac{7}{1.75}=4\)
• For a general \(x\), \(y = 1.75x\)

For Table E (\(y = 10.75x\)):

• When \(x = 1\), \(y = 10.75\)
• When \(y = 129\), \(x=\frac{129}{10.75}=12\)
• For a general \(x\), \(y = 10.75x\)

For Scenario D:

If \(g = 5\) gallons, \(t=3\times5 = 15\) hours

For Scenario A:

The number of tiles \(y = 10.75\times28.5=306.375\)