To finish the demonstration that the quadrilateral JKLM is a rhombus we need to prove that side JK is congruent with side LM.
The length of a segment with endpoints (x1, y1) and (x2, y2) is calculated as follows:
Substituting with points L(1,6) and M(4,2) we get:
Given that opposite sides are parallel, all sides have the same length, and, from the diagram, the quadrilateral is not a square, we conclude that it is a rhombus.
34/2380
we can divide by 2 on top and bottom
17/1190
17 is prime so that is simplest form
nevermind
1/70
if you wanted decimal, use yor calculator and get 0.01428571428571428571428571428571
Answer:
Step-by-step explanation:
well i need to know how much you are buying to know the total price XD
9514 1404 393
Answer:
- -√5
- 3/5
- -4/5
Step-by-step explanation:
The relevant relations are ...
sec = ±√(tan² +1)
cos = 1/sec
csc = 1/sin = ±1/√(1 -cos²)
Sine and Cosecant are positive in quadrants I and II. Cosine and Secant are positive in quadrants I and IV.
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1. sec(θ) = -√((-2)² +1) = -√5
2. cos(θ) = 1/sec(θ) = 1/(5/3) = 3/5
3. csc(θ) = -1/√(1 -(-3/5)²) = -√(16/25) = -4/5