Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
45 = 3x3x5
Or you could simplify it to 3^2x5
Answer:
c
Step-by-step explanation:
So we are given two points that the line crosses, the origin and (9, -3), we can calculate the slope m of the line with these data, dividing the y segment by the x segment:
m = (-3 - 0)/(9 - 0) = -3/9
m = -1/3
then we can use the point slope line equation to find the line equation, lets use the point origin (0,0) to do so:
y - y1 = m(x - x1), where x1, y1 are the coordinates of a point that the line crosses:
y - 0 = (-1/3)(x - 0)
y = <span>(-1/3)x
so this is the equation of the line, slope -1/3 and y intercept 0</span>
Answer:
x^3 + 1/x^3 = 488
Step-by-step explanation:
- x^2 + 1/x^2 = 62
- x^2 + 1/x^2 + 2 = 64
- ( adding 2 in both sides )
- (x + 1/x ) ^2 = 64
- x + 1/x = 8
now,
- ( x+ 1/x ) ^ 3 = 512
- x^3 + 1/x^3 + 3 × x × 1/x ( x + 1/x )
- x^3 + 1/x^3 + 3 ( 8 )
- ( since x + 1/x = 8 )
- x^3 + 1/x^3 + 24 = 512
- x^3 + 1/x^3 = 488
hence, we got x^3 + 1/x^3 = 488
