The function of the area of the square is A(t)=121
Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area
Lets assume the length of side of square is x
11 
⇒x=11t
Area of square=
Area of square=
{as the length of side is 11t}{varies by time}
Area of square=121
Therefore,The function of the area of the square is A(t)=121
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Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area
Lets assume the length of side of square is x
11 
⇒x=11t
Area of square=
Area of square=
{as the length of side is 11t}{varies by time}
Area of square=121
Therefore,The function of the area of the square is A(t)=121
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Answer:
12 and 13
Step-by-step explanation:
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<span>$20 per hour
</span><span>8 hrs a day
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It will have a vertical asymptote when the denominator approaches zero.
2x-6=0
2x=6
x=3
So the vertical asymptote is the vertical line x=3