Answer:
2) 40 and 41
3) L=23 and W=11
4) W≥10 while L≥14
Explanation:
2)
The sum of 2 consecutive integers is less than 83.
Find a pair of integers with the greatest sum.
Let n be an integer, then (n+1) is the next integer.
So we have the following inequality from the first line:
n+(n+1)<83
n+n+1<83
2n+1<83
Subtract 1 on both sides:
2n<82
Divide both sides by 2:
n=41
n+1=41+1=42.
So 41+42 is equal to 83.
We want it to be less than 83.
So the pair 40 and 41 would work
40+41 would be the next greater sum and it satisfies the restriction of 2 consecutive pair of integers having sum less than 83.
3)
The length of a rectangle is 12 meters longer than it's width.
L=12+W
The perimeter is 68. A rectangles opposite sides are congruent and there are 4 sides. The perimeter is the sum of the side measurements:
L+L+W+W=68
Combine l ike terms:
2L+2W=68
Plug 1st equation into the perimeter equation:
2(12+W)+2W=68
Distribute:
24+2W+2W=68
Combine like terms:
24+4W=68
Subtract 24 on both sides:
4W=44
Divide both sides by 4:
W=11
L=12+W=12+11=23.
4) The length of a rectangle is 4 cm more than the width: L=4+W.
The perimeter is at least 48 cm: 2L+2W≥48. (At least mean equals to or greater than.)
Plug the first equation into the inequality there:
2(4+W)+2W≥48.
Distribute:
8+2W+2W≥48
Combine like terms:
8+4W≥48
Subtract 8 on both sides:
4W≥40
Divide both sides by 4:
W≥10
L=4+W
Solving this for L gives us L-4=W (I subtracted 4 on both sides).
So if W≥10 and W=L-4 then:
L-4≥10
Add 4 on both sides:
L≥14.
So while the width is greater than or equal to 10, the length is 14 or greater.