Answer:
hi
Step-by-step explanation:
hi
(2+4+6)*8 hopefully that's what you are looking for
Answer:

Step-by-step explanation:
<u>Quadratic Function</u>
Standard Form of Quadratic Function
The standard representation of a quadratic function is:

where a,b, and c are constants.
When the zeros of f (x1 and x2) are given, it can be written as:
f(x)=a(x-x1)(x-x2)
Where a is a constant called the leading coefficient.
We are given the two roots of f: x1=-3 and x2=4, thus:
f(x)=a(x+3)(x-4)
We also know that f(5)=8, thus:
f(5)=a(5+3)(5-4)=8
Operating:
a(8)(1)=8
Solving:
a=1
The function is:
f(x)=1(x+3)(x-4)
Operating:

Answer:
Option a is right
Step-by-step explanation:
Given that as part of a research project on student debt at TWU, a researcher interviewed a sample of 35 students that were chosen at random concerning their monthly credit card balance.
Sample average = 2573
Variance = 4252
Sample size = 35
STd deviation of X = 
Score of student selected at random X=1700
Corresponding Z score = 
Rounding of we get Z score = -13.4
option a is right
Basically, what this asks you is to maximize the are A=ab where a and b are the sides of the recatangular area (b is the long side opposite to the river, a is the short side that also is the common fence of both corrals). Your maximization is constrained by the length of the fence, so you have to maximize subject to 3a+b=450 (drawing a sketch helps - again, b is the longer side opposite to the river, a are the three smaller parts restricting the corrals)
3a+b = 450
b = 450 - 3a
so the maximization max(ab) becomes
max(a(450-3a)=max(450a-3a^2)
Since this is in one variable, we can just take the derivative and set it equal to zero:
450-6a=0
6a=450
a=75
Plugging back into b=450-3a yields
b=450-3*75
b=450-225
b=215
Hope that helps!