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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]
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