Your question is incomplete.
If you want me to tell the ratio, its 3:2
Rectangle and square
However, I would need to know which figures are involved seeing as there are no pictures of anything to work with.
For A divide 2 and 90 for B put 1:and whatever you came up with A. And finnaly for C put 1 x your constant of p
Yes. Take for example a square and an ellipsis with the same perimeter. The family of ellipses with the same perimeter can have any area between that of a circle to zero (if it is extremely “thin” i.e. if its eccentricity is large). The circle has the maximum area of any other shape with the same perimeter, so the square has the same area of one of the intermediate ellipses.
Answer:
1. y=(x+3)^3. Zero: x=-3 multiplicity 3.
2. y=(x-2)^2 (x-1). Zeros: x=2 multiplicity 2; x=1 multiplicity 1.
3. y=(2x+3)(x-1)^2. Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.
Step-by-step explanation:
1. y=(x+3)^3
![y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3](https://tex.z-dn.net/?f=y%3D0%5C%5C%20%28x%2B3%29%5E3%3D0%5C%5C%20%5Csqrt%5B3%5D%7B%28x%2B3%29%5E3%7D%3D%5Csqrt%5B3%5D%7B0%7D%5C%5C%20x%2B3%3D0%5C%5C%20x%2B3-3%3D0-3%5C%5C%20x%3D-3)
Zero: x=-3 multiplicity 3.
2. y=(x-2)^2 (x-1)
Zeros: x=2 multiplicity 2; x=1 multiplicity 1
3. y=(2x+3)(x-1)^2
Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.
