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Sagot :

Answer:

( 6(x)² + 12(x) - 90 ) ft³

Explanation:

  • deep end water: 2(x)² + 12(x) + 10
  • shallow end water: 4(x)² - 100

addition problem: 2(x)² + 12(x) + 10 + 4(x)² - 100

total volume of water:

  • 2(x)² + 12(x) + 10 + 4(x)² - 100
  • 2(x)² + 4(x)² + 12(x) + 10 - 100
  • 6(x)² + 12(x) - 90

Answer:

a)  [tex]\sf total \ volume \ of \ pool= (2x^2+12x+10)+(4x^2-100) \ ft^3[/tex]

b)  [tex]\sf total \ volume \ of \ pool =(6x^2+12x-90) \ ft^3[/tex]

Step-by-step explanation:

Given:

  • volume of water in the deep end:  [tex]\sf (2x^2+12x+10) \ ft^3[/tex]
  • volume of water in the shallow end:  [tex]\sf (4x^2-100) \ ft^3[/tex]

a)  Total volume of water = deep end volume + shallow end volume

[tex]\implies \sf total= (2x^2+12x+10)+(4x^2-100)[/tex]

b) Simplify:

[tex]\sf \implies 2x^2+12x+10+4x^2-100[/tex]

Collect like terms:

[tex]\sf \implies 2x^2+4x^2+12x+10-100[/tex]

Combine like terms:

[tex]\sf \implies 6x^2+12x-90[/tex]