Answer:
- time = 1second
- maximum height = 16m
Step-by-step explanation:
Given the height of a pumpkin t seconds after it is launched from a catapult modelled by the equation
f(t)=-16t²+32t... (1)
The pumpkin reaches its maximum height when the velocity is zero.
Velocity = {d(f(x)}/dt = -32t+32
Since v = 0m/s (at maximum height)
-32t+32 = 0
-32t = -32
t = -32/-32
t = 1sec
The pumpkin reaches its maximum height after 1second.
Maximum height of the pumpkin is gotten by substituting t = 1sec into equation (1)
f(1) = -16(1)²+32(1)
f(1) = -16+32
f(1) = 16m
The maximum height of the pumpkin is 16m
Answer:
first we need to substitute g(t) value in h(t) value and then evaluate the expression as follows :
h(g(t))=3(t-1)²+4(t-1).
by solving
h(g(t))=3(t²+1-2t)+4t-4
h(g(t))=3t²-2t-1 will.be the final answer
4x + 3y = 12
3y = -4x + 12
y = -4/3x + 4........so the slope (or gradient) is -4/3...because in y = mx + b form, the slope(gradient) is in the m position and the y int is in the b position....so if u wanted to know the y axis, it would be (0,4)
the x intercept (where the line crosses the x axis) can be found by subbing in 0 for y in the original equation or the slope intercept equation, and solving for x.
4x + 3(0) = 12
4x = 12
x = 12/4 = 3....so the x intercept is (3,0)
One angle is inside a right triangle
Answer:
<em>(x - 2)^2 + (y + 1)^2 = 26</em>
Step-by-step explanation:
A circle with center O(2, -1) that passes through the point A(3, 4).
=> The radius of this circle is OA which could be calculated by:
OA = sqrt[(3 - 2)^2 + (4 - (-1))^2] = sqrt[1^2 + 5^2] = sqrt[26]
The equation of a circle with center O(a, b) and radius r could be written as:
(x - a)^2 + (y - b)^2 = r^2
=> The equation of circle O above with center O(2, -1) and radius = sqrt(26) is shown as:
(x - 2)^2 + (y - (-1))^2 = (sqrt(26))^2
<=>(x - 2)^2 + (y + 1)^2 = 26
Hope this helps!
