The function below describes the number of students who enrolled at a university, where f(t) represents the number of students a
1 answer:
Answer:
18,500 , 1 , 1.03
Step-by-step explanation:
I think your question is missed of key information, allow me to add in and hope it will fit the original one.
f(t) =18,500(1.03)^t
1st blank:
19,055
1.03
3
18,500
2nd blank:
t years
2 years
3 years
1 year
3rd blank:
19,055
3
18,500
1.03
My answer:
Given that: f(t) =18,500*
1st blank:
Initial value when t = 0, so we have:
f(0) =18,500*
So we choose D for 1st blank
2nd and 3rd blank:
Because it's an exponential function f(t) =18,500* , with every value of t increment f(t) increase by a factor of 1.03 . So Every 1 year the number of students who enroll at the university increases by a factor of 1.03
=> 2nd blank: 1
=> 3rd blank: 1.03
Hope it will find you well.
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Ooh, fun
what I would do is to make it a piecewise function where the absolute value becomse 0
because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up
so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points
we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5
A.
B.
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the last one you can do yourself, it is
the sum is
so the area under the curve is
what I would do is to make it a piecewise function where the absolute value becomse 0
because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up
so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points
we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5
A.
B.
sepearte the integrals
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