<span> 303.6 that is your anser</span>
Answer:
20 meters
Step-by-step explanation:
The track is circular so it means that after Patrick raced the entire track he is back at the starting point. In other words, every 440 meters he is back to the beginning.
So we would have that, if he races round the track twice, he would run 440(2) = 880 meters and he would be back at the starting point.
The problem asks us how far is he from the starting point at the 900 meter mark. If at 880 meters he is at the starting point, then at 900 meters he would be 
meters from the starting point.
Answer:
?
Step-by-step explanation:
You can’t know that, because you don’t know how many total apples there are.
Circumfrence: pi times diameter
7.9*
circumfrence is approximatley 24.806
area:
a=
*(1/2(d))^2
a= 48.99
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
