Answer:
Distance of JK = 15 unit
Step-by-step explanation:
Given:
J(4,8)
K(-1,-2)
Find:
Distance of JK
Computation:
Distance = √(x₂-x₁)² + (y₂-y₁)²
Distance of JK = √(-1-4)² + (-2-8)²
Distance of JK = √25 + 100
Distance of JK = √125
Distance of JK = 15 unit
⓵
-4ỿ = 8
Simplify the left side in order to isolate the ỿ!
-4ỿ = 8
+4 +4
Ỿ = 12
⓶
× + 3y - 3z = -26
Simplify the left side in order to isolate the ×, ỿ and z!
× + 3y - 3z = -26
÷3 ÷3
× + ỿ - 3z = -8,66 periodic
÷-3 ÷-3
× + ỿ + z = 2,88 periodic
⓷
2× - 5ỿ + z = 19
Simplify the left side in order to isolate the ×, ỿ and z!
2× - 5ỿ + z = 19
÷2 ÷2
× - 5ỿ + z = 9,5
÷-5 ÷-5
× + ỿ + z = -1,9
F(x)=3 domain: negative infinity, positive infinity range: 3
g(x)=x+7 domain: - infinity, + infinity range: - infinity, + infinity
12k - 2k +16 you add the 2 and 10 on the first half and the 3 and 13 on the other half
