First set of data:
Mean - 6.5
Absolute deviation - 2.4
Second set of data:
Mean - 4.475
Absolute deviation - 2.275
Given:
The given quadratic polynomial is :

To find:
The quadratic polynomial whose zeroes are negatives of the zeroes of the given polynomial.
Solution:
We have,

Equate the polynomial with 0 to find the zeroes.

Splitting the middle term, we get




The zeroes of the given polynomial are -3 and 4.
The zeroes of a quadratic polynomial are negatives of the zeroes of the given polynomial. So, the zeroes of the required polynomial are 3 and -4.
A quadratic polynomial is defined as:




Therefore, the required polynomial is
.
A(t)=A0*(1/2)^(t/30)
where t=elapsed years
A0=initial mass at t=0.
Answer:
73
Step-by-step explanation:
Since DA is the diameter, this means that you can divide the circle into two half-circles. You will solve with the right half-circle. A half-circle has a 180 total degrees. Add up the values given and solve for x.
(2x) + (7x - 4) + (85) = 180
2x + 7x - 4 + 85 = 180
9x - 4 + 85 = 180
9x + 81 = 180
9x = 99
x = 11
Now that you have the value of x, solve for m∠CZB.
m∠CZB = 7x - 4
m∠CZB = 7(11) - 4
m∠CZB = 77 - 4
m∠CZB = 73
Answer:
The answer is actually 74 days. I know how to algebraically solve this, however I am unsure how to transpose my work onto the computer.
Step-by-step explanation:
I can also tell you that edgenuity will say 74 is correct.