It is in the air 5.625 seconds.
Using the quadratic formula,
0 is when the ball is first launched, 5.625 is when it hits the ground again.
Answer:
m < EAD = 29 degrees
m < CAB = 119 degrees
Given :
The question states that m < CAE = m<FAB = 61 degrees and m<DAF = 90 degrees
Solution:
1. Since line CAF and EAB intersect each other, m<CAF = m< EAF - (opposite vertical angles are equivalent)
2. m<BAC + m<EAC = 180 degrees (sum of linear pair)
3. m<CAB = 180 degrees - m<EAC
4. Equation 1: m<CAB = m<EAF = 119 degrees
5. m<EAF = m<EAD + m< DAF
6. m<EAD = m<EAF - m<DAF
7. m<EAD = 119 degrees-90 degrees = 29 degrees
Hope this helps!!! :)
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Answer:y = 3x + 2
Step-by-step explanation: y = mx + b
Where m is the slope and b is the y-intercept.