2.5 hours i think but i might be wrong :)
Answer:
x= 16±√-64 over 2
Step-by-step explanation:
x= -(-16)±√(-16)²-4×1×80 over 2×1
x= 16±√256-320 over 2
x= 16±√-64 over 2
can also be represented with all real numbers.
The given equations are

(1)

(2)
When t=0, obtain

Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means

.
Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1
y'(0) = 1/2.
Here, y' means

.
Because

, obtain

Answer:
The slope of the curve at t=0 is 1/6.
Answer:
y= -3/5x + 2
Step-by-step explanation:
It’s A I’m pretty sure it’s right