Answer:
23.55
Step-by-step explanation:
The answer is actually a repeating decimal with an infinite amount of 5s but you only really need to show 2 or 3 to show that it's repeating
Answer:
ASA and AAS
Step-by-step explanation:
We do not know if these are right triangles; therefore we cannot use HL to prove congruence.
We do not have 2 or 3 sides marked congruent; therefore we cannot use SSS or SAS to prove congruence.
We are given that EF is parallel to HJ. This makes EJ a transversal. This also means that ∠HJG and ∠GEF are alternate interior angles and are therefore congruent. We also know that ∠EGF and ∠HGJ are vertical angles and are congruent. This gives us two angles and a non-included side, which is the AAS congruence theorem.
Since EF and HJ are parallel and EJ is a transversal, ∠JHG and ∠EFG are alternate interior angles and are congruent. Again we have that ∠EGF and ∠HGJ are vertical angles and are congruent; this gives us two angles and an included side, which is the ASA congruence theorem.
Answer:
o (16, 8, 12)
Step-by-step explanation:
1st number: 8*2 =16
2nd number:12-4=8
3rd number: given
Answer:
see below
Step-by-step explanation:
72pi cubic inches
1/3 pi r ^2 * h = 72 pi
1/3 r^2 h = 72
You can theoretically find infinite answers to this and get enough points for the semester where you don't have to do work anymore. I will list a couple more possible answers below
r = radius, h = height
r = 2, h = 54
r = 3, h = 24
r = 4, h = 27/2
r = √(1/pi), h = 72pi
Answer:
The given point is a solution to the given system of inequalities.
Step-by-step explanation:
Again, we can substitute the coordinates of the given point into the system of inequalities. We know that the x-coordinate and y-coordinate of 
are 
and 
, respectively.
Plugging these values into the first inequality, 
, gives us 
, which simplifies to 
. This is a true statement, so the given point satisfies the first inequality. We still need to check if it satisfies the second inequality though, because if it doesn't, it won't be a solution to the system.
Plugging the coordinates into the second inequality, 
, gives us 
, which simplifies to 
. This is also a true statement, so the given point satisfies the second inequality as well. Therefore, 
is a solution to the given system of inequalities since it satisfies all of the inequalities in the system. Hope this helps!
