Use the points (1997, 13) and (2005, 22) to write the slope intercept form of an equation for the line of fit shown in the scatt
er plot. Round values to the nearest hundredth.
A. y=0.89x+1985.44
B. y=1.13x-2243.65
C. y=1.13x-2233.63
D. y=0.89x-1985.44
A. y=0.89x+1985.44
B. y=1.13x-2243.65
C. y=1.13x-2233.63
D. y=0.89x-1985.44
1 answer:
You might be interested in
Answer:
The perimeter of the kite JKLM is 50 units ⇒ C
Step-by-step explanation:
The longest diagonal of the kite is its axis of symmetry which means the longest diagonal divides the kite into two congruent triangles
∵ JKLM is a kite
∵ KM is the longest diagonal
∴ KM is divides the kite into two congruent triangles
∴ Δ JKM is congruent to Δ LKM
∴ JK = LK
∴ JM = ML
∵ ML = 10 units
∴ JM = 10 units
∵ The diagonals of the kite are perpendicular
∴ KM ⊥ JL
∵ KM and JL intersected at N
∴ m∠JNK = 90°
∵ KM is the axis of symmetry of kite JKLM
∴ KM bisects JL at N
∴ JN = NL = JL
∵ JL = 18 units
∴ JN = (18)
∴ JN = 9 units
In Δ JNK
∵ m∠JNK = 90°
∵ JN = 9 units
∵ NK = 12 units
- By using Pythagoras Theorem
∵ (JK)² = (JN)² + (NK)²
∴ (JK)² = (9)² + (12)²
∴ (JK)² = 81 + 144
∴ (JK)² = 225
- Take √ for both sides
∴ JK = 15 units
∵ Jk = LK
∴ Lk = 15 units
∵ The perimeter of the kite is the sum of its 4 sides
∴ P = JK + KL + LM + MJ
∵ JK = LK = 15
∵ JM = ML = 10
∴ P = 15 + 15 + 10 + 10
∴ P = 50
The perimeter of the kite JKLM is 50 units