It is true that the product of two consecutive even integers are always one less than the square of their average.
<u>Step-by-step explanation</u>:
Let the two consecutive odd integers be 1 and 3.
- The product of 1 and 3 is (1
3)=3 - The average of 1 and 3 is (1+3)/2 =4/2 = 2
- The square of their average is (2)² = 4
∴ The product 3 is one less than the square of their average 4.
Let the two consecutive even integers be 2 and 4.
- The product of 2 and 4 is (24)=8
- The average of 2 and 4 is (2+4)/2 =6/2 = 3
- The square of their average is (3)² = 9
∴ The product 8 is one less than the square of their average 9.
Thus, It is true that the product of two consecutive even integers are always one less than the square of their average.
Answer:
We can have two cases.
A quadratic function where the leading coefficient is larger than zero, in this case the arms of the graph will open up, and it will continue forever, so the maximum in this case is infinite.
A quadratic function where the leading coefficient is negative. In this case the arms of the graph will open down, then the maximum of the quadratic function coincides with the vertex of the function.
Where for a generic function:
y(x) = a*x^2 + b*x + c
The vertex is at:
x = -a/2b
and the maximum value is:
y(-a/2b)
Answer:
Step-by-step explanation:
Let x, y , and z be the numbers.
Then the geometric sequence is 
Recall that term of a geometric sequence are generally in the form:
This implies that:
a=32 and 
Substitute a=32 and solve for r.
Take the fourth root to get:
![r=\sqrt[4]{\frac{81}{256} }](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B4%5D%7B%5Cfrac%7B81%7D%7B256%7D%20%7D)
Therefore 
The scale factor is 1/2 or a half
