Answer:
The maximum height of the projectile is 90 ft
Step-by-step explanation:
Here, we want to get the maximum height reached by the projectile
The answer here will be the y-coordinate value of the vertex form of the given equation
so firstly, we have to write the equation in the vertex form
We have this as;
y = -16t^2 + 64t + 26
That will be;
y = a(x-h)^2 + k
y = -16(x-2)^2 + 90
where the vertex of the equation is;
(-h,k)
K
in this case is 90 and thus, that is the maximum height of the projectile
3.4*10^-4
hope this helps!!
-3(b - 7)
Multiply -3 into the parenthesis.
= -3b + 21.
Answer:
xy+2x+3y+6=xy-5 .............(1)
xy-3x+2y-6=xy.....,.,(2)
2x+3y=-11
-4x+2y=6
first equation multiply by 2
(2)*(2x+3y)=-11
4x+6y=-22
-4x+2y=6
8y=-16 y=(-2),,,,, x=(-2,5)
The standard equation of a circle is
(x-h)^2 + (y-k)^2 = r^2
where the center is at point (h,k)
From the statement of the problem, it is already established that h = 2 and k = -5. What we have to determine is the value of r. This could be calculated by calculating the distance between the center and point (-2,10). The formula would be
r = square root [(x1-x2)^2 + (y1-y2)^2)]
r = square root [(2--2)^2 + (-5-10)^2)]
r = square root (241)
r^2 = 241
Thus, the equation of the circle is