Answer:
Step-by-step explanation:
Question
Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work.
perimeter of a triangle = AB+AC+BC
Using the distance formula
AB = sqrt(-2-5)²+(2-2)²
AB = sqrt(-7)²
AB =sqrt(49)
AB =7
BC = sqrt(-2+2)²+(2-5)²
BC = sqrt(0+3²)
BC =sqrt(9)
BC =3
AC= sqrt(-2-5)²+(2-5)²
AC= sqrt(-7)²+3²
AC =sqrt(49+9)
AC =sqrt58
Perimeter = 10+sqrt58
Answer:
16t + 10
Explanation:
Step 1 - Add like terms
-16t + 32t + 10
16t + 10
1) 9m
2) 1/64
3) 50d^3
4) pq^2 (p-q) (p-q)
= pq^2 ( p^2 - 9p -9p +18)
= pq^4 - 18p^2q^2 + 8pq^2
= q^2 - 18p + 8
Answer:
It should be 5 but my brain is literally dead
Step-by-step explanation:
Please give brainlest
