Answer:
19. d. 0.5 m
20. a. 1.5 m
Step-by-step explanation:
19.
Let the dimensions of the rectangular bench be l, w, and h.
[tex]\rm V = lwh[/tex]
Given:
- [tex]\rm V = \frac{3}{8} \ m^3[/tex]
[tex]\implies \rm \frac{3}{8} = (w+1)(w)(w)[/tex]
[tex]\implies \rm \frac{3}{8} = (w^2 + w)(w)[/tex]
[tex]\implies \rm \frac{3}{8} = w^3 + w^2[/tex]
[tex]\implies \rm 8(\frac{3}{8} ) = 8(w^3 + w^2)[/tex]
[tex]\implies \rm 3 = 8w^3 + w^2[/tex]
[tex]\implies \rm 8w^3 + w^2 - 3 = 0[/tex]
[tex]\implies \rm (2w - 1)(4w^2+6w+3)=0[/tex]
[tex]\therefore \rm (2w - 1) = 0 \ or \ (4w^2 + 6w + 3) = 0[/tex]
[tex]\rm If \ (2w-1)=0, \ then \ w = \frac{1}{2}[/tex]
[tex]\rm If \ (4w^2 + 6w + 3) = 0, \ then \ the \ width \ is \ negative, \ which \ is\ unrealistic[/tex]
Therefore the width of the rectangular bench is d. 0.5 m
20.
[tex]\rm l = w + 1[/tex]
Substitute the value of width.
[tex]\rm l = 0.5 +1[/tex]
[tex]\boxed{\rm l = \rm1.5}[/tex]
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