Answer:
x = 15 / 7
Step-by-step explanation:
<h3>
Answer: C) 4</h3>
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Work Shown:
Expand out the left hand side
(x-k)(x-5)
x(x-5)-k(x-5)
x^2-5x-kx+5k
Note that the constant term here is 5k. Yes k seems like a variable, but it's actually a constant. Once we know k, we can replace it to get a fixed number. In this case, k = 4 since 5k = 5*4 = 20 to have it match with the 20 at the end of x^2-9x+20
For the x terms, we have -5x-kx = -5x-4x = -9x which matches with the middle term of x^2-9x+20
Therefore,
(x-k)(x-5) = x^2-9x+20
updates to
(x-4)(x-5) = x^2-9x+20
The -4 and -5 multiply to 20, and they also add to -9. This helps confirm we have the right k value.
<h3>
Answer: (1,3)</h3>
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Explanation:
Note that the two shaded regions overlap to form the darker shaded region. The point (1,1) is only in the green region, but not in the blue region. It needs to be in both regions for it to be a solution to the system. In contrast, the point (1,3) is on the boundary of the blue region and in the green region as well. It's a solution point because of this.
Points on the boundary are included because of the "or equal to" as part of the inequality sign. Hence the solid boundary lines rather than dashed boundary lines.
The solution (1,3) means that x = 1 and y = 3 pair up together. It says that Cody can buy x = 1 hotdog and y = 3 packs of peanuts so that he has at least 4 snacks, and he stays within the $7 budget.
Going back to (x,y) = (1,1) for a moment, this isn't a solution because x+y = 1+1 = 2 is not 4 or larger. In other words, he only buys 2 snacks here instead of 4 or more.
<span>These are points where f ' = 0. Use the quiotent rule to find f '.
f ' (x) = [(x^3+2)(1) - (x)(3x^2)] / (x^3+2)^2
f ' (x) = (2 - 2x^3) / (x^3 + 2)^2
Set f ' (x) = 0 and solve for x.
f ' (x) = 0 = (2-2x^3) / (x^3+2)^2
Multiply both sides by (x^3+2)^2
(x^3+2)^2 * 0 = (x^3+2)^2 * [(2-2x^3)/(x^3+2)^2]
0 = 2 - 2x^3
Add 2x^3 to both sides
2x^3 + 0 = 2x^3 + 2 - 2x^3
2x^3 = 2
Divide both sides by 2
2x^3 / 2 = 2 / 2
x^3 = 1
Take cube roots of both sides
cube root (x^3) = cube root (1)
x = 1. This is our critical point
2) Points where f ' does not exist.
We know f ' (x) = (2-2x^3) / (x^3+2)^2
You cannot divide by 0 ever so f ' does not exist where the denominator equals 0
(x^3 + 2)^2 = 0. Take square roots of both sides
sqrt((x^3+2)^2) = sqrt(0)
x^3 + 2 = 0. Add -2 to both sides.
-2 + x^3 + 2 = -2 + 0
x^3 = -2. Take cube roots of both sides.
cube root (x^3) = cube root (-2)
x = cube root (-2). This is where f ' doesnt exist. However, it is not in our interval [0,2]. Thus, we can ignore this point.
3) End points of the domain.
The domain was clearly stated as [0, 2]. The end points are 0 and 2.
Therefore, our only options are: 0, 1, 2.
Check the intervals
[0, 1] and [1, 2]. Pick an x value in each interval and determine its sign.
In [0, 1]. Check 1/2. f ' (1/2) = (7/4) / (17/8)^2 which is positive.
In [1, 2]. Check 3/2. f ' (3/2) = (-19/4) / (43/8)^2 which is negative.
Therefore, f is increasing on [0, 1] and decreasing on [1, 2] and 1 is a local maximum.
f (0) = 0
f (1) = 1/3
f (2) = 1/5
Therefore, 0 is a local and absoulte minimum. 1 is a local and absolute
maximum. Finally, 2 is a local minimum. </span><span>Thunderclan89</span>
