Quadratic formula: (-b +/- sqrt(b^2 - 4ac)) / 2a
a = 10
b = -1
c = 9
1 +/- sqrt((-1)^2 - 4(10)(9)) / 2(10)
1 +/- sqrt(1 - 360) / 20
x = 1 +/- sqrt(359i) / 20
Hope this helps!
68.80 because 7 is greater than 5
It is given that, B ≅ BC and AD ≅ CD
We need BD perpendicular to AC, then only we can say triangles AXB and CXB are congruent using the HL theorem.
If BD perpendicular to AC, means that AB and CB are the hypotenuse of triangles AXB and CXB respectively.
from the given information ABCD is a square
If BD and AC bisect each other then AX = CX
Then only we can immediately possible to prove that triangles AXD and CXD are congruent by SSS congruence theorem
Answer:
t = 4 seconds
Step-by-step explanation:
The height of the projectile after it is launched is given by the function :
t is time in seconds
We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0
So,
The above is a quadratic equation. It can be solved by the formula as follows :
Here, a = -16, b = 32 and c = 128
Neglecting negative value, the projectile will land after 4 seconds.
