Answer:
1.
<u>An extraneous solution is a root of a transformed equation that is not a root of the original equation as it was excluded from the domain of the original equation.</u>
It emerges from the process of solving the problem as a equation.
2.I begin like:
The vertical asymptotes will occur at those values of x for which the denominator is equal to zero:
for example:
x² − 4=0
x²= 4
doing square root on both side
x = ±2
Thus, the graph will have vertical asymptotes at x = 2 and x = −2.
To find the horizontal asymptote, the degree of the numerator is one and the degree of the denominator is two.
<h3>
Answer:</h3><h3>If you know one ratio in a proportion , you can use that information to find values in the other equivalent ratio. </h3>
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
Answer: (21,12)
Step-by-step explanation:
