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Below is the solution:
Backing out the bonus, we get 2711 - 250 = 2461, which is 161 more than his last check
<span>So 161 / 2300 = 0.07, or 7%
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F(6)= -10 because -2(6) is -12. -12+2 = -10
g(4)= 44 because pemdas mean we do exponents first so -4(-4)= 16. 16x2= 32. -4(-3)=12. 32+12=44
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: m<PQT=60 and m<PQS=4x+14
m<PQS and m<SQT are same
m<PQT = m<PQS + m<SQT = m<PQS + m<PQS = 2m<PQS
m<PQT = 2m<PQS
So m<PQS =
Plug in the given information and solve for x
4x + 14 =
4x + 14 = 30 (subtract 14 on both sides)
4x = 16 (Divide by 4 on both sides)
x = 4
The value of x is 4
