Answer:
C. 4x^3 - 14x^2y + 14xy^2 - 4y^3
Step-by-step explanation:
Given:
Product of 2x^2 – 3xy + y^2 and 2x – 4y
Product means multiplication
(2x^2 – 3xy + y^2) (2x – 4y)
Open the bracket
= 4x^3 - 8x^2y - 6x^2y + 12xy^2 + 2xy^2 - 4y^3
Simplify the like terms
= 4x^3 - 14x^2y + 14xy^2 - 4y^3
The answer is
C. 4x^3 - 14x^2y + 14xy^2 - 4y^3
The correct statement regarding the local minimum of the function is:
Over the interval [4, 7], the local minimum is -7.
<h3>What is the missing information?</h3>
The graph of the function is missing, and is given at the end of the answer.
<h3>What is a local minimum in a function f(x)?</h3>
A local minimum in a function f(x) is a value of x at which the function changes from decreasing to increasing. The minimum is the value of y at this point.
The graph of the function shows that:
- The function is increasing until x = 0, when y = 26.
- Then, the function decreases until x = 5.1, when y = -7.
- For x > 5.1, the function increases to infinity.
From the description of the graph, the function changes from decreasing to increasing at point (5.1, -7), hence this point is a local minimum and the correct statement is given by:
Over the interval [4, 7], the local minimum is -7.
More can be learned about local minimums at brainly.com/question/2437551
#SPJ1
Y=1x+1?? that’s what u get when u use the omni calculator
![f(n)\in\mathcal O(g(n))](https://tex.z-dn.net/?f=f%28n%29%5Cin%5Cmathcal%20O%28g%28n%29%29)
is to say
![|f(n)|\le M_1|g(n)|](https://tex.z-dn.net/?f=%7Cf%28n%29%7C%5Cle%20M_1%7Cg%28n%29%7C)
for all
![n](https://tex.z-dn.net/?f=n)
beyond some fixed
![n_1](https://tex.z-dn.net/?f=n_1)
.
Similarly,
![d(n)\in\mathcal O(h(n))](https://tex.z-dn.net/?f=d%28n%29%5Cin%5Cmathcal%20O%28h%28n%29%29)
is to say
![|d(n)|\le M_2|h(n)|](https://tex.z-dn.net/?f=%7Cd%28n%29%7C%5Cle%20M_2%7Ch%28n%29%7C)
for all
![n\ge n_2](https://tex.z-dn.net/?f=n%5Cge%20n_2)
.
From this we can gather that
![|f(n)+d(n)|\le|f(n)|+|d(n)|\le M_1|g(n)|+M_2|h(n)|\le M(|g(n)|+|h(n)|)](https://tex.z-dn.net/?f=%7Cf%28n%29%2Bd%28n%29%7C%5Cle%7Cf%28n%29%7C%2B%7Cd%28n%29%7C%5Cle%20M_1%7Cg%28n%29%7C%2BM_2%7Ch%28n%29%7C%5Cle%20M%28%7Cg%28n%29%7C%2B%7Ch%28n%29%7C%29)
where
![M](https://tex.z-dn.net/?f=M)
is the larger of the two values
![M_1](https://tex.z-dn.net/?f=M_1)
and
![M_2](https://tex.z-dn.net/?f=M_2)
, or
![M=\max\{M_1,M_2\}](https://tex.z-dn.net/?f=M%3D%5Cmax%5C%7BM_1%2CM_2%5C%7D)
. Then the last term is bounded above by
![M(|g(n)|+|h(n)|)\le2M\max\{|g(n)|,|h(n)|\}](https://tex.z-dn.net/?f=M%28%7Cg%28n%29%7C%2B%7Ch%28n%29%7C%29%5Cle2M%5Cmax%5C%7B%7Cg%28n%29%7C%2C%7Ch%28n%29%7C%5C%7D)
from which it follows that
Answer:
d
Step-by-step explanation: