Answer is a, 57.49.
17.42-12.60-9.62= $-4.80+62.29=57.49
Answer:
In words the answer is between t=0 and t=2.
In interval notation the answer is (0,2)
In inequality notation the answer is 0<t<2
Big note: You should make sure the function I use what you meant.
Step-by-step explanation:
I hope the function is h(t)=-16t^2+32t because that is how I'm going to interpret it.
So if we can find when the ball is on the ground or has hit the ground (this is when h=0) then we can find when it is in the air which is between those 2 numbers.
0=-16t^2+32t
0=-16t(t-2)
So at t=0 and t=2
So the ball is in the air between t=0 and t=2
Interval notation (0,2)
Inequality notation 0<t<2
principal (p)=62500,Time (T)=1.5 years,Rate (R)=8% Ammount=p (1+R÷100) =62500 × 1.1664 =72900 again, compound interest = p(1+R÷100)-1=62500×0.1664 = 10400
I think a non- integer rational number but I’m not real sure on that.
Let us model this problem with a polynomial function.
Let x = day number (1,2,3,4, ...)
Let y = number of creatures colled on day x.
Because we have 5 data points, we shall use a 4th order polynomial of the form
y = a₁x⁴ + a₂x³ + a₃x² + a₄x + a₅
Substitute x=1,2, ..., 5 into y(x) to obtain the matrix equation
| 1 1 1 1 1 | | a₁ | | 42 |
| 2⁴ 2³ 2² 2¹ 2⁰ | | a₂ | | 26 |
| 3⁴ 3³ 3² 3¹ 3⁰ | | a₃ | = | 61 |
| 4⁴ 4³ 4² 4¹ 4⁰ | | a₄ | | 65 |
| 5⁴ 5³ 5² 5¹ 5⁰ | | a₅ | | 56 |
When this matrix equation is solved in the calculator, we obtain
a₁ = 4.1667
a₂ = -55.3333
a₃ = 253.3333
a₄ = -451.1667
a₅ = 291.0000
Test the solution.
y(1) = 42
y(2) = 26
y(3) = 61
y(4) = 65
y(5) = 56
The average for 5 days is (42+26+61+65+56)/5 = 50.
If Kathy collected 53 creatures instead of 56 on day 5, the average becomes
(42+26+61+65+53)/5 = 49.4.
Now predict values for days 5,7,8.
y(6) = 152
y(7) = 571
y(8) = 1631
