13.
(a)
![y=-5x^2+21x-3](https://tex.z-dn.net/?f=y%3D-5x%5E2%2B21x-3)
for x = 2:
![y=-5(2)^2+21(2)-3=-20+42-3=19](https://tex.z-dn.net/?f=y%3D-5%282%29%5E2%2B21%282%29-3%3D-20%2B42-3%3D19)
for x = 5
![y=-5(5)^2+21(5)-3=-125+105-3=-23](https://tex.z-dn.net/?f=y%3D-5%285%29%5E2%2B21%285%29-3%3D-125%2B105-3%3D-23)
(b)
The graph is:
(c)
Using the graph:
![\begin{gathered} x=1.8,y=18.6 \\ x=0.548,y=7 \\ x=3.65,y=7 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D1.8%2Cy%3D18.6%20%5C%5C%20x%3D0.548%2Cy%3D7%20%5C%5C%20x%3D3.65%2Cy%3D7%20%5Cend%7Bgathered%7D)
![\begin{gathered} x=1.8 \\ y=-5(1.8)^2+21(1.8)-3=-16.2+37.8-3=18.6 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D1.8%20%5C%5C%20y%3D-5%281.8%29%5E2%2B21%281.8%29-3%3D-16.2%2B37.8-3%3D18.6%20%5Cend%7Bgathered%7D)
![\begin{gathered} y=7 \\ -5x^2+21x-3=7 \\ -5x^2+21x-10=0 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D7%20%5C%5C%20-5x%5E2%2B21x-3%3D7%20%5C%5C%20-5x%5E2%2B21x-10%3D0%20%5Cend%7Bgathered%7D)
Using the quadratic formula:
Check out the attached image for the answers.
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Statement 2 is blank, but it has the reasoning "Corresponding angles postulate"
Because BD || AE, we know that the corresponding angles are congruent.
One pair of corresponding angles is angle 1 and angle 4. This is because they are on the same side of the transversal AC and they are both above their parallel line counter-part. Similarly, angle 2 and angle 3 are another corresponding pair.
So you'll have "angle1=angle4, angle3=angle2" in the first blank slot
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Reason 3 is blank. The statement is that triangle ACE is similar to triangle BCD. The reason why the are similar is the AA (angle angle) similarity postulate. This says that if you know two pairs of angles are congruent, then the triangles are similar. The two pairs of angles were mentioned back on the previous line (line 2)
So you'll put "Angle-Angle Similarity Postulate" in the second blank.
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Look at the line just above the last line. Here we have
1 + (BA/CB) = 1 + (DE/CD)
If we subtract 1 from both sides, we end up with,
BA/CB = DE/CD
which is what will go in the last blank space
Side Note: The last statement will always be what you want to prove. So you can just look at the very top of the problem where it says "Prove:" under the "Given" part. Then just copy/paste the statement you want to prove, which in this case is BA/CB = DE/CD
Answer: The maximum volume of the package is obtained with a cross section of side 18 inches and a length of 36 inches.
Step-by-step explanation: This is a optimization with restrictions problem.
The restriction is that the perimeter of the square cross section plus the length is equal to 108 inches (as we will maximize the volume, we wil use the maximum of length and cross section perimeter).
This restriction can be expressed as:
being x: the side of the square of the cross section and L: length of the package.
The volume, that we want to maximize, is:
If we express L in function of x using the restriction equation, we get:
We replace L in the volume formula and we get
To maximize the volume we derive and equal to 0
We can replace x to calculate L:
The maximum volume of the package is obtained with a cross section of side 18 inches and a length of 36 inches.
Answer:
This can have multiple solutions. Let's use the equation m + p = 15.
Step-by-step explanation: