let
(hypotenuse ) c = 10 in
(perpendicular) a = 8 in
(base) b =?
By using Pythagoras theorem
c^2= a^2+b^2
(10)^2 = (8)^2 +(b)^2
100 = 64 +(b)^2
100- 64 = (b)^2
36 = (b)^2
√36 = b
6 = b
hence b= 6 in
Well there really wouldn't be a combo since everything is different so I'm saying nun? that's what I'm going with. ( If this is wrong correct me )
The 3 bubble is the correct answer.
being that you're looking for 3.60 in quarters(25) and nickles (5) which add up to be 3.60 so 25q+5n=360
Step-by-step answer:
This is a regular heptagon, means it has 7 <em>congruent</em> sides and 7 <em>congruent </em>vertex angles.
To work with polygons, there is a very important piece of information that you must know to solve the majority of related problems.
This is:
sum of exterior angles of polygons = 360 degrees.
If you don't remember the 360 degrees, think of the sum of exterior angles of an equilateral triangle, which is 3*(180-60)=3*120=360! It works!
For a regular heptagon, c = 360/7=51.43 degrees approx.
This means that each vertex angle measures
vertex angle = 180-c
So since 2d+the vertex angle = 360, we have
2d+(180-c)=360
solve for d:
2d=360-(180-c)=180+c
d=(180+c)/2=90+c/2=115.71 degrees. (approx.)
