Answer:
x = 143
Step-by-step explanation:
The chord- chord angle x is half the sum of the measures of the arcs intercepted by the angle and its vertical angle, that is
x = (140 + 146) = 0.5 × 286 = 143
Can some one help me answer? i dont understand
1 answer:
Based on the question when the ball is thrown vertically upwards by Melvin who stands at the edge of a cliff is modelled by an equation with variables t (representing the time in seconds) and h (representing the height in metres); h = t ( 8 - 5t )
a) After one second since the ball was released or in essence when t = 1, the height of the ball can be calculated by substituting for t in the equation thus solving for h.
when t = 1 ; h = ( 1 ) [ 8 - 5 ( 1 )]
h = 3
∴ at one second, the height of the ball was 3 metres
b) (i) when t = 5; h = ( 5 ) [ 8 - 5 ( 5 )]
h = - 85
b) (ii) after exactly 5 seconds after releasing, the height of the ball was -85 metres.
This suggests that the ball went upwards vertically, reached its highest point, and then fell downwards, passing the edge of the cliff where Marvin stood and goes down below the edge of the cliff (where it was released) by 85 metres, hence the - 85 metres.
c) assume that the top of the cliff is represented by h = 0; that is when the height of the ball is at the edge of the cliff, thus not above or below but at the height of 0.
when h = 0; (0) = t ( 8 - 5t)
0 = 8t - 5t²
by using the quadratic formula x = [ -b ± √(b² - 4ac) ] ÷ 2a where a = -5 b = 8 c = 0
Now t = [ -8 ± √(8² - 4(-5)(0)) ] ÷ 2(-5)
t = [ -8 ± √(64) ] ÷ (- 10)
⇒ t = (-8 + 8) ÷ (- 10) or t = (-8 - 8) ÷ (-10)
⇒ t = 0 & t = 1.6
∴ the ball has two points at which it is position is at the edge (when h = 0); this occurs when time is 0 seconds but also when time is 1.6 seconds. This coincides with part b when it was established that the path of the ball is that of a parabola with a maximum point.
a) After one second since the ball was released or in essence when t = 1, the height of the ball can be calculated by substituting for t in the equation thus solving for h.
when t = 1 ; h = ( 1 ) [ 8 - 5 ( 1 )]
h = 3
∴ at one second, the height of the ball was 3 metres
b) (i) when t = 5; h = ( 5 ) [ 8 - 5 ( 5 )]
h = - 85
b) (ii) after exactly 5 seconds after releasing, the height of the ball was -85 metres.
This suggests that the ball went upwards vertically, reached its highest point, and then fell downwards, passing the edge of the cliff where Marvin stood and goes down below the edge of the cliff (where it was released) by 85 metres, hence the - 85 metres.
c) assume that the top of the cliff is represented by h = 0; that is when the height of the ball is at the edge of the cliff, thus not above or below but at the height of 0.
when h = 0; (0) = t ( 8 - 5t)
0 = 8t - 5t²
by using the quadratic formula x = [ -b ± √(b² - 4ac) ] ÷ 2a where a = -5 b = 8 c = 0
Now t = [ -8 ± √(8² - 4(-5)(0)) ] ÷ 2(-5)
t = [ -8 ± √(64) ] ÷ (- 10)
⇒ t = (-8 + 8) ÷ (- 10) or t = (-8 - 8) ÷ (-10)
⇒ t = 0 & t = 1.6
∴ the ball has two points at which it is position is at the edge (when h = 0); this occurs when time is 0 seconds but also when time is 1.6 seconds. This coincides with part b when it was established that the path of the ball is that of a parabola with a maximum point.
You might be interested in