For the first one you would do:
X rad 3=7 divide by rad 3 in both sides and you don’t wanna leave a radical in the denominator so multiply rad 3 in both denom. And numerator and then you get 7rad3/3.
A 30:60:90 triangle always have n:2n:n rad 3 so since 2n=10 you would get n=5 and since BC is n rad3 so you would get 5rad 3.
In a 45:45:90 triangle it is n:n:n rad 2, since AC=3 then AB= 3 and then BC= 3 rad 2
Since AC and AB are congruent you they both would be 12 so it is true.
Answer:
See proof below
Step-by-step explanation:
We will use properties of inequalities during the proof.
Let 
. then we have that 
. Hence, it makes sense to define the positive number delta as 
(the inequality guarantees that these numbers are positive).
Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that 
, and if we prove this, we are done. To prove it, let 
, then 
. First, 
then 
hence 
On the other hand, 
then 
hence 
. Combining the inequalities, we have that 
, therefore as required.
Answer:
for n ≤ 3
we are going to put a circle on the 3 and make it filled in then you are going to start an arrow starting from the 3 going to the left all the way to the -10.
Answer:
131˙
Step-by-step explanation:
Supplementary angles are angles that add up to 180˙. If you subtract 49 from 180, you get 131.
Brainliest? Maybe?
