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.
There are 2 ways you can write this.
You can either do /1 or /100

The most used is
Hey there!
The opposite of the square root of 25 is -5, and this is an <u>integer and a rational number and a real number.</u>
<u></u>
Hope it helps and have a great day!
Answer:
sinB= 3/5, tanA=4/3, cosB=4/5
Step-by-step explanation:
The y-intercepts are
(0,0) and (5,0)
The equation of the quadratic function is
f(t) = at(t-5)
Solving for a
12 = a(3)(3-5)
a = -2
The equation of the quadractic function is
f(t) = -2t(t -5)
f(t) = -2t² + 10t<span />