Step-by-step explanation:
after I answered the other one(s), you should be really able to do this yourself.
it is precisely the same method and formula just with different numbers.
what's the problem now ?
y = -16x² + 247x + 141
we assume, the ground is at 0 ft.
so, we need to solve
0 = -16x² + 247 x + 141
the general solution for such quadratic equations is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = -16
b = 247
c = 141
x = (-247 ± sqrt(247² - 4×-16×141))/(2×-16) =
= (-247 ± sqrt(61009 + 9024))/-32 =
= (-247 ± sqrt(70033))/-32
x1 = (-247 + 264.6374879...)/-32 = -0.551171497... s
x2 = (-247 - 264.6374879...)/-32 = 15.9886715... s
again, the negative solution for time did not make any sense, so, x2 is our solution.
the rocket will hit the ground after about 15.99 seconds.
If it's vertical, it's called undefined.
Formula:
Hope this helped, if not, don’t spam.
Answer:
9
Step-by-step explanation:
6a=54
a=54/6
a=9
Given that: Triangle ABC is translated to triangle A'B'C' by the transformation
The points are A(4, -4), B(2, -5), C(0, 6), D(-2, 5).
To find: Coordinates of A'.
The coordinates of A are (4,-4).
Using the given rule of transformation, we get
On simplification, we get
Therefore, the coordinates of point A' are (6,-9).
