Answer:
Suppose that you have a square of side length x, the area of this square will be:
A = x^2.
Now, by dynamics, we know that the position of an object that is falling down from a height H, can be written as:
H(t) = (-g/2)*t^2 + H.
We can see a pattern, x^2 is used in both.
Now, profit can be also modeled with quadratic equations, where our objective is to find the maximum of the quadratic (so we can have the maximum profit)
Then the parent function that is useful for gravity, calculating area and profit is the quadratic function:
f(x) = x^2
Answer: A=100(12)t5.27
Step-by-step explanation:
Assume / and ^ got lost from eqns.
A(t) = initial × (1/2)^(t/halflife)
So initial = 100, halflife = 5.27,
A=100(12)t5.27 means A=100(1/2)^(t/5.27)
so that A is 50 when t is 5.27
Answer:
x = 2.67°
GKJ = 40°
Step-by-step explanation:
Angle HKJ = 6x + 4
Angle HKG = 9x - 4
Segment bisects Angle GKJ, it means that angle HKJ is the same as angle HKG.
You have the measure of both angles in terms of x.
Angle HKJ = Angle HKG
6x + 4 = 9x - 4
4 + 4 = 9x - 6x
8 = 3x
2.67 = x
Angle HKJ:
6(2.67) + 4 = 20°
Angle HKG:
9(2.67) - 4 = 20°
Angle GKJ = HKJ + HKG = 20 + 20 = 40°
The changing rate of the amount of water in the watering can is a constant rate of 0.15 gallon per minute.
The initial volume of water in the can at time = 0 is 5 gallons
The function will be a linear function in the form of
where
is the volume of water in the can after
minutes
is the changing rate
is the initial volume when time is 0
Substitute all the values into the form we have
Answer:0-9
24
10-19
20
20-29
32
Step-by-step explanation:
