The answer is B(11, 2sqrt(12) )
proof
the main equation of the circle is (x-x1)²+(y-y1)²=R²
where C(x1, y1) is the center
so if the center is the origin, it is O(0,0), and the equation becomes
<span> (x)²+(y)²=R²
</span>and the circle passes through the point (-5,2) so we can write
(-5)²+12²=R², it implies R= sqrt(25+144)=sqrt(169)=13
and for <span>B(11, 2sqrt(12) ) </span>11²+ (2sqrt(12))²= 121 + 48= 169= 13
it is checked.
Answer:
(A) 5.47 km
(B) 1.38 km
Step-by-step explanation:
The relationship between Pressure (mmHg) and Height (km) is given by the equation below:
p =760 e⁻ 0.145h----------------------------------------------------- (1)
(a) The height of the aircraft can be calculated by making 'h' the subject of equation (1)
In (p/760) = -0.145 h
h = -In (p/760) /0.145
h= In (760/p)/ 0.145------------------------------------------------- (2)
For the aircraft, p =344 mmHg; substituting into (2)
h = In (760/344)/0.145
= 0.7926/0.145
= 5.4662 km
≈ 5.47 km
(b) p= 622 mmHg at the mountain top. Substituting into equation (2) we have:
h = In (760/622)/0.145
= 0.2003 / 0.145
= 1.3813 km
≈ 1.38 km
Answer:
a) about 0.7 seconds to 5.1 seconds.
b) Listed below.
Step-by-step explanation:
h - 1 = -5x^2 + 29x
h = -5x^2 + 29x + 1
a) We will find the amount of time it takes to get to 18 meters.
18 = -5x^2 + 29x + 1
-5x^2 + 29x + 1 = 18
-5x^2 + 29x - 17 = 0
We will then use the quadratic formula to find the answer.
[please ignore the A-hat; that is a bug]
= 
= 
= 
= 
and 
= 0.6616970714 and 5.138302929
So, the time period for which the baseball is higher than 18 metres ranges from about 0.7 seconds to 5.1 seconds.
b) Restrictions on the domain and range of the function are that the domain and range can never be negative, since time cannot be negative, and height cannot be negative. The height cannot exceed the vertex of the parabola, since that is the highest the ball will ever go. It cannot exceed that height since gravity will cause the ball to fall down.
Hope this helps!
