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
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Answer:
the teachers would have 24 students per teacher. 10 students per tutor. There would need to be 11 tutors
Step-by-step explanation:
each tutor would have 10 students then if you do 11 times 10 you get 11
(hope this helps can I pls have brainlist (crown)☺️)
Answer:
$103.88
Step-by-step explanation:
$98 + 6% is 103.88.
Performing the indicated multiplication on the left side gives you:
8x-24=8x-24,
8x=8x
1=1
These expressions are true for any real value of x and thus has infinitely many solutions...
Easy peasy
just subsitute I(x) for the x in the h(x) so
h(I(s))=-(2s+3)^2-4
distribute and simplify
h(I(s))=-(4s^2+12s+9)-4
h(I(s))=-4s^2-12s-9-4
h(I(s))=-4s^2-12s-13