So, the best way to do this is translate it to clockwise. 90 degrees counterclockwise is equal to 270 degrees clockwise. So, basically, to rotate, you would follow the following format for each point-
(X,Y) -> (-Y,X)
Now, you do it for each of the points.
A= (-5,5), so A' would be (-5,-5)
B= (-1,5), so B' would be (-5,-1)
C= (-5,4), so C' would be (-4,-5)
D= (-1,4) so D' would be (-4,-1)
Notice, how all the points end up in the square below it. Each quadrant has a specific number. The top right is quadrant 1, the top left is quadrant 2, the bottom left is quadrant 3, and the bottom right is quadrant 4. If you are rotating 270 degrees clockwise, you move to the right, like a clock. That puts the new rectangle in quadrant 3. That is a way to check your work.
Now, just so you know for future reference, the following are also different formats for different problems--
A 90 degree Clockwise rotation about the origin will be (X,Y) -> (Y, -X) *Note, -x just stands for the opposite. Say your original x is a negative number. Then the prime (new) x will be positive.
A 180 degree Clockwise rotation about the origin would be (X,Y) -> (-X,-Y) *Note, -y also stands for the opposite.
A 270 degree clockwise rotation about the origin would be (X,Y) -> (-Y,X).
For translating---
90 degrees Clockwise = 270 degrees Counter
270 degrees Clockwise = 90 degrees Counter
Hope this helped!
Check the picture below. So,the parabola looks like so, notice the distance "p". since the parabola is opening to the right, then "p" is positive, thus is 5.

Answer: 3, 4,5 and 17, 15,8
give the measures of the legs and hypotenuse of a right triangle.
Step-by-step explanation:
In order for the measures of the legs and hypotenuse given to form a right angle triangle, they must be Pythagorean triples. A Pythagoras triple is a set of numbers that perfectly satisfy the Pythagorean theorem. The Pythagorean theorem is expressed as
Hypotenuse² = opposite side² + adjacent side². We will apply the theorem to each set of numbers given.
1) 3, 4, 5
5² = 3² + 4² = 9 + 16
25 = 25
It is a Pythagorean triple
2) 4, 11, 14
14² = 11² + 4² = 121 + 16
196 = 137
It is a Pythagorean triple
3) 9, 14, 17
17² = 14² + 9² = 196 + 81
289 = 277
It is not a Pythagorean triple
4) 8, 14, 16
16² = 14² + 8² = 196 + 64
256 = 260
It is not a Pythagorean triple
5) 8, 15 , 17
17² = 15² + 8²
289 = 225 + 64
289 = 289
It is a Pythagorean triple
Therefore, 3, 4,5 and 17, 15,8
give the measures of the legs and hypotenuse of a right triangle.
The pool can hold 65.84 ft³ of water
<u>Explanation:</u>
Given:
Shape of pool = octagonal
Base area of the pool = 22 ft²
Depth of the pool = 3 feet
Volume, V = ?
We know:
Area of octagon = 2 ( 1 + √2) a²
22 ft² = 2 ( 1 + √2 ) a²

a² = 
a² = 4.55
a = 2.132 ft
Side length of the octagon is 2.132 ft
We know:
Volume of octagon = 

Therefore, the pool can hold 65.84 ft³ of water