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
Learn more about 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
Learn more about area here:
brainly.com/question/27683633
#SPJ4
<span>Assuming that the red and blonde hair are considered the 3/8ths of the school with light hair, 270 children in the school have red or blonde hair.</span>
Y = mx + c
-4 = 2(3) + c
-4 = 6 + c
-10 = c
y = mx + c
y = 2x - 10
Hence, the equation is y=2x-10.
(1) For the parabola on the bottom row, the domain would be R and the range would be y ≥ -5
(2) For the hyperbola on the bottom row, the domain would be R\{3} (since there is an asymptote at x = 3) and the range would be R\{4} (since there is an asymptote at y = 4)
(3) For the square root function on the bottom row, the domain would be x ≥ -5 and the range would be (-∞, -2]
(4) For the function to the very right on the bottom row, the domain would be R and the range would be (-∞, -3]
Answer:ok
Step-by-step explanation: