Answer:
31.9secs
6,183.3m
Step-by-step explanation:
Given the equation that models the height expressed as;
h(t ) = -4.9t²+313t+269
At the the max g=height, the velocity is zero
dh/dt = 0
dh/dt = -9,8t+313
0 = -9.8t + 313
9.8t = 313
t = 313/9.8
t = 31.94secs
Hence it takes the rocket 31.9secs to reach the max height
Get the max height
Recall that h(t ) = -4.9t²+313t+269
h(31.9) = -4.9(31.9)²+313(31.9)+269
h(31.9) = -4,070.44+9,984.7+269
h(31.9) = 6,183.3m
Hence the maximum height reached is 6,183.3m
Area=1/2 times bas time height
given that area=10
height=5
find base
10=1/2 times base times 5
times 2 both siddes
20=base times 5
divide oth sides by 5
4=base
the base is 4m
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Equation of the line in slope intercept form is
y = 1/4 * x + 5. Slope is 1/4. Intercept is 5.
If x= 4, y = (1/4)*4 + 5 = 6.
The point (4,6) lies on the line.
Equation of the line in point slope form is
(y - y1)= m * (x - x1) where m is slope & (x1,y1) is point on the line.
Substituting slope m= 1/4 & point (4,6),
Equation of the line in point-slope form is:
(y - 6) = 1/4 * (x - 4).
Sanjay C.
