Answer:
You will need 20 sides to complete the loop.
Step-by-step explanation:
The question isn't quite clear given how small the corner is, but I assume that we are looking to complete the circle if the pentagon and square are repeated in a loop
We can also see - assuming that those are proper equal-sided polygons, that PQ is the same length as PV
With that in mind, We can solve this by noting that the angle of a corner in a square is 90 degrees, and in a pentagon it's 108 degrees.
108 - 90 is equal to 18. This means that PQ is at eighteen degrees to YP. Also, QM, (which will be equivalent to the next VP is eighteen degrees to PQ.
This means that each polygon is rotated 18 degrees relative to it's neighbour.
With all that we can say that the total polygons we need to form a circle is 360/18 = 20, So you will need 20 polygons, or ten squares and ten pentagons to complete the loop.
Answer:
https://www.mathpapa.com/
Step-by-step explanation:
Problem
For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground
Solution
We know that the x coordinate of a quadratic function is given by:
Vx= -b/2a
And the y coordinate correspond to the maximum value of y.
Then the best options are C and D but the best option is:
D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a
The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.
Answer:

Step-by-step explanation:
Connect points I and K, K and M, M and I.
1. Find the area of triangles IJK, KLM and MNI:

2. Note that

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.
I’m not sure but you can use a calculator I saw a fantasy one at Walmart with A SCREEN ON IT!??? it’s $100 but it would probably help ALOT