What is the image of E for a 120° counterclockwise rotation about the center of the regular hexagon?

N

Step-by-step explanation:

Since this hexagon has six sides, and to end up back at E again we’d need to rotate 360°, this means for every side we cross it’s a 60° turn.

Therefore, ° would be two sides.

Since we are moving counter-clockwise, the image for E would be to the left of it.

Two sides/points to the left of E is N, so the image of E for a 120° counterclockwise rotation around the center of this shape would be N.

Hope this helped!

something that characterises a right triangle is the fact that we may use pythagoras' theorem to find any one of its sides given that we have the length of the other two sides. here, we have the length of the hypotenuse, √(87) yd, and one other side, √(23) yd, thus we can easily find the length of the remaining side using pythagoras' theorem as defined below:

c^2 = a^2 + b^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

thus, given that c = √(87) and a = √(23), we may substitute these into the formula we defined above to get:

c^2 = a^2 + b^2

(√87)^2 = (√23)^2 + b^2 (substitute c = √87 and a = √23 into the equation)

87 = 23 + b^2 (evaluate (√87)^2 = 87 and (√23)^2 = 23 - remember that the square root of a number, squared will equal that very same number)

64 = b^2 (subtract 23 from both sides)

√64 = b (take the square root of both sides)

8 = b (evaluate √64 = 8)

therefor, the length of the unknown leg in the right triangle is 8 yd (first answer).