Question

1. the ordered pair (10¢, 25¢) shows the value of one dime and one quarter. which group of ordered pairs shows the values of the coins when there are 2, 3, 4, and 5 of each coin?

a (20¢, 25¢), (30¢, 50¢), (40¢, 75¢), (50¢, $1.00)
b (20¢, 50¢), (40¢, 75¢), (40¢, $1.00), (50¢, $1.25)
c (10¢, 50¢), (20¢, 75¢), (30¢, $1.00), (40¢, $1.25)
d (20¢, 50¢), (30¢, 75¢), (40¢, $1.00), (50¢, $1.25)

1. in 2 hours, a mechanic can change the oil in 5 cars. which set of ordered pairs represents the hours the mechanic spends to change oil in cars?

a (2, 5)(4, 10)(6, 15)(8, 20)
b (2, 5)(4, 7)(6, 9)(8, 11)
c (2, 5)(4, 10)(8, 15)(16, 20)
d (2, 4)(6, 8)(5, 10)(15, 20)

1. allan made a pattern using the rule “add 4,” and jim made a pattern using the rule “subtract 5.” they both started at 30. if allan’s number is x and jim’s number is y, which list of ordered pairs represents the first four numbers in their patterns?

a (30, 30), (35, 26), (40, 22), (45, 18)
b (30, 30), (34, 25), (38, 20), (42, 15)
c (30, 30), (26, 35), (22, 40), (18, 45)
d (30, 30), (25, 34), (20, 38), (15, 42)

Explanation:

Question 31

Step1: Calculate value of 2 dimes and 2 quarters

Value of 2 dimes: \(2\times10\phi = 20\phi\)
Value of 2 quarters: \(2\times25\phi=50\phi\)
Ordered - pair: \((20\phi,50\phi)\)

Step2: Calculate value of 3 dimes and 3 quarters

Value of 3 dimes: \(3\times10\phi = 30\phi\)
Value of 3 quarters: \(3\times25\phi = 75\phi\)
Ordered - pair: \((30\phi,75\phi)\)

Step3: Calculate value of 4 dimes and 4 quarters

Value of 4 dimes: \(4\times10\phi=40\phi\)
Value of 4 quarters: \(4\times25\phi = 100\phi=\$1.00\)
Ordered - pair: \((40\phi,\$1.00)\)

Step4: Calculate value of 5 dimes and 5 quarters

Value of 5 dimes: \(5\times10\phi = 50\phi\)
Value of 5 quarters: \(5\times25\phi=125\phi=\$1.25\)
Ordered - pair: \((50\phi,\$1.25)\)

Step1: Find the rate

The mechanic changes the oil in 5 cars in 2 hours. The rate is \(\frac{5}{2}\) cars per hour.

Step2: For \(t = 2\) hours, \(n = 5\) cars (given)

For \(t = 4\) hours, \(n=\frac{5}{2}\times4 = 10\) cars
For \(t = 6\) hours, \(n=\frac{5}{2}\times6=15\) cars (not in the options)
For \(t = 4\) hours, if we consider the pattern of doubling the time and doubling the number of cars (since \(2\) hours - \(5\) cars, \(4\) hours - \(10\) cars (\(5\times2\)), \(8\) hours - \(20\) cars (\(5\times4\))). But if we assume a linear relationship \(n=\frac{5}{2}t\), when \(t = 2\), \(n = 5\); when \(t=4\), \(n = 10\); when \(t = 8\), \(n = 20\) (Option A: \((2,5),(4,10),(6,15),(8,20)\) has a correct linear relationship with a slope of \(\frac{5}{2}\))

Step1: Find Allan's numbers

Allan's rule: \(x_{n}=x_{n - 1}+4\), starting at \(x_1 = 30\)
\(x_1=30\), \(x_2=30 + 4=34\), \(x_3=34+4 = 38\), \(x_4=38 + 4=42\)

Step2: Find Jim's numbers

Jim's rule: \(y_{n}=y_{n - 1}-5\), starting at \(y_1 = 30\)
\(y_1=30\), \(y_2=30-5 = 25\), \(y_3=25-5=20\), \(y_4=20-5 = 15\)

Answer:

D. \((20\phi,50\phi),(30\phi,75\phi),(40\phi,\$1.00),(50\phi,\$1.25)\)

Question 32