Question

A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second? [hint:if x is the length of the shortest board, then x,(x+3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x+3) + 2x ≤ 91 and 2x≥(x+3)+5] 1. 5 ≤ x ≤ 10 cm2. 8 ≤ x ≤ 22 cm3. 8 ≤ x ≤ 15 cm4. 5 ≤ x ≤ 15 cm

Answer 1

Let the length of the shortest piece, The length of the second price is (x+3)(x+3)cm. The length of the third piece is 2x cm, Total length of the three prieces must be ≤91cm≤91cm.

xcm+(x+3)cm+2xcm≤91cmxcm+(x+3)cm+2xcm≤91cm

=>4x+3≤91=>4x+3≤91

Subtracting 3 from number 4 on both sides, 4x4≤8844x4≤884

x≤22x≤22----(1)

Step 2:

Also third piece is at least 5 cm longer than the second piece. therefore

2x≥(x+3)+52x≥(x+3)+5

=>2x≥x+8=>2x≥x+8

Subtracting x from both sides,

2x−x≥82x−x≥8

x≥8x≥8 -----(2)

Step 3:

From (1) and (2)

8≤x≤228≤x≤22

The possible length of the shortest side must be greater than or equal to 8 cm and less than or equal to 22 cm.

Please mark brainliest

Answer 2

Answer:

Step-by-step explanation:

Let the of the shortest board be x cm.

lenght of the second board

= 3 cm longer than the shortest side

= x+3 cm

& lenght of the third board

= twice the shortest lenght

= 2x cm

x+( x+3)+2x<= 91

4x+3<= 91

4x<=88

x<=88/4

×<=22

2x>=(x+3)+5

2x>=×+8

2×-×>=8

x>=8.