Ineed the explanation to how to solve these questions like how to set up the equation or inequality.

2) 40 and 41

3) L=23 and W=11

4) W≥10 while L≥14

Step-by-step 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.

he will have 300 because he is really broke and 300 is in his imangine

step-by-step explanation: he ehe heh step by the thy

i dont know if you cant find a way its best to actually add all of them start from 105 tho

step-by-step explanation: