For a), this is clearly a given as it is literally to the right of where it says “Given:”
For b), since ON bisects ∠JOH, this means that it splits it into two separate angles - JON and HON, which are similar due to that bisects mean that it splits it equally into two halves
For c), since NO is the same thing as NO, it is equal to itself
For d), since AAS (angle-angle-side) congruence states that if there are two angles that are congruent (proved in a) and b) ) as well as that a side is congruent (proved in c) ), two triangles are congruent
For e), since two triangles are congruent, every side must have one side that it matches up to in the other triangle. As the opposite side of angle H is JO and the opposite side of angle J is OH, and ∠J=∠H, those two are congruent. As JN and HN are the two sides left, they must be congruent.
Feel free to ask further questions!
Part A:
There are 8 possible outcomes.
HHH
TTT
HTH
THT
TTH
HHT
THH
HTT
Part B:
There is 2 outcomes that consist of all head or all tails.
Part C:
You would expect about 6 students to get all head or all tails.
Hope this helps c:
The answer is going to be 7/8. hope that helped
It's a cube so x should also equal 6
Answer:
The length and width of another rectangular field with same perimeter but a larger area is 80 m by 70 m
Step-by-step explanation:
The perimeter of the existing field is
2(l + b)
= 2(90 + 60) = 2(150) = 300 yards
So we want another field having the same perimeter but a larger area
The area we have here is 90 * 60 = 5,400 square yards
If we had 80 by 70
Perimeter will still be 2(70 + 80) = 150
But the area will be 80 * 70 = 5,600 square yards
