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
<u>Given</u>:
Line m is parallel to line n.
The measure of ∠1 is (4x + 15)°
The measure of ∠2 is (9x + 35)°
We need to determine the measure of ∠1
<u>Value of x:</u>
From the figure, it is obvious that ∠1 and ∠2 are linear pairs.
Thus, we have;

Substituting the measures of ∠1 and ∠2, we get;




Thus, the value of x is 10.
<u>Measure of ∠1:</u>
The measure of ∠1 can be determined by substituting x = 10 in the measure of ∠1
Thus, we have;



Thus, the measure of ∠1 is 55°
Forget the '34' part of 8:34 and the '18' minutes from 3 hours and 18 minutes for now.
Just add 3 hours on to 8, to get 11. Then, bring back the 34 and 18 and add them together. 34 + 12 = 52. Now put that 52 on the end of the 11, and you get 11:52 as the end time.