Perimeter of rectangle = length + length + width + width
To find the combinations, think of two numbers that each multiplied by 2 and added up to give 12 or 14
Rectangle with perimeter 12
Say we take length = 2 and width = 3
Multiply the length by 2 = 2 × 2 = 4
Multiply the width by 3 = 2 × 3 = 6
Then add the answers = 4 + 6 = 10
This doesn't give us perimeter of 12 so we can't have the combination of length = 2 and width = 3
Take length = 4 and width = 2
Perimeter = 4+4+2+2 = 12
This is the first combination we can have
Take length = 5 and width = 1
Perimeter = 5+5+1+1 = 12
This is the second combination we can have
The question doesn't specify whether or not we are limited to use only integers, but if it is, we can only have two combinations of length and width that give perimeter of 12
length = 4 and width = 2
length = 5 and width = 1
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Rectangle with perimeter of 14
Length = 4 and width = 3
Perimeter = 4+4+3+3 = 14
Length = 5 and width = 2
Perimeter = 5+5+2+2 = 14
Length = 6 and width = 1
Perimeter = 6+6+1+1 = 14
We can have 3 different combinations of length and width
Answer:
The second choice is the same so it is the last one
Answer:
Advance tickets cost $30; same-day tickets cost $35.
Step-by-step explanation:
Let a = the cost of an advance ticket
and s = the cost of a same-day ticket
We have two conditions:
(1) a + s = 65
(2) 15a + 20s = 1150
Subtract a from each side of (1) (3) s = 65 - a
Substitute (3) into (2) 15a + 20(65 - a) = 1150
Distribute the 20 15a + 1300 - 20a = 1150
Combine like terms 1300 - 5a = 1150
Subtract 1300 from each side -5a = -150
Divide each side by -5 (4) a = 30
Substitute (4) into (1) 30 + s = 65
Subtract 30 from each side s = 35
Advance tickets cost $30; same-day tickets cost $35.
Check:
(1) 30 + 35 = 65 (2) 15 × 30 + 20 × 35 = 1150
65 = 65 450 + 700 = 1150
1150 = 1150
A. Range would be your answer
