Answer:

Step-by-step explanation:
step 1
Find the area of the circular region
The area of the circle is equal to

we have

assume

substitute


step 2
Find the number of people who live in the region
Multiply the area by the density
so

Round to the nearest ten thousand

Not really well frased question. im guessing you mean the hypotonouse is 61 and the height is 11. you can use the pythagorean theorem by doing a squared plus b squared equals c squared. you know 2 variables, a and c. so you can do 11 squared plus b squared equals 61 squared. so 121 plus b squared equals 3721. then 3721-121 is 3600. the square root of 3600 is 60, giving 60 as the answer.
360 tiles because you have to multiply 12x30.
Answer:
x=-1
Step-by-step explanation:
The lines stand for absolute value ( how far a number is from 0), so each number should be a positive, then you just solve for x. Isolate x, then divide on both sides, and you should get -x=1, which is equal to x=-1.
C.AAS
Because vertical angles are congruent.