So for this one, all you have to do is plot the points it's giving you :)
Then you connect the lines and you get a shape from A-b-c-d
Answer:
True, 1, 3, 5
Step-by-step explanation:
1. Angle L is sircumsribed angle, thenMJ⊥JL and MK⊥KL, so the measure of the angle L is
360°-131°-90°-90°=49° (the sum of all interior angles in the quadrilateral is 360°)
True
2. False, because m∠L=49°.
3. If the measure of arc HJ is 98°, then the measure of the central angle HMJ is 98°.
True
4. If the measure of arc HJ is 98°, then the measure of the central angle HMJ is 98° and the meausre of inscribed angle HKJ subtended on the arc HJ is half of the central angle measure and is equal to 49°.
False
5. If the measure of arc HJ is 98°, then the measure of the arc HK is
360°-131°-98°=131°
True
Answer:
(7,4)
Step-by-step explanation:
midpoint formula (x1+x2)/2 , (y1+y2)/2
Look at x and y for the endpoints separately
midpoint x value = ( 1 + x)/2
4= (1 + x)/2
8 = 1 + x
x = 7
midpoint y value = (2 + y)/2
3 = (2 + y)/2
6 = 2 + y
y = 4
endpoint (x,y) = (7,4)
Answer:
1) 1 element
2) 13 elements
3) 22 elements
4) 40 elements
Step-by-step explanation:
1) Only one element will have no tails: the event that all the coins are heads.
2) 13 elements will have exactly one tile. Basically you have one element in each position that you can put a tail in.
3) There are 
elements that have exactly 2 tails. From those elements we have to remove the only element that starts and ends with a tail and in the middle it has heads only and the elements that starts and ends with a head and in the 11 remaining coins there are exactly 2 tails. For the last case, there are 
possibilities, thus, the total amount of elements with one tile in the border and another one in the middle is 78-55-1 = 22
4) We can have:
- A pair at the start/end and another tail in the middle (this includes a triple at the start/end)
- One tail at the start/end and a pair in the middle (with heads next to the tail at the start/end)
For the first possibility there are 2 * 11 = 22 possibilities (first decide if the pair starts or ends and then select the remaining tail)
For the second possibility, we have 2*9 = 18 possibilities (first, select if there is a tail at the end or at the start, then put a head next to it and on the other extreme, for the remaining 10 coins, there are 9 possibilities to select 2 cosecutive ones to be tails).
This gives us a total of 18+22 = 40 possibilities.
