2.) Mila completed a half marathon race over the weekend. The full race was
1 answer:
You might be interested in
Answer:
(a). y'(1)=0 and y'(2) = 3
(b).
(c).
Step-by-step explanation:
(a). Let the curve is,
So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value which lies in the domain of f where the derivative is 0.
Therefore, y'(1)=0
Also given that the derivative of the function y(t) is 3 at t = 2.
Therefore, y'(2) = 3.
(b).
Given function,
Differentiating the above equation with respect to x, we get
Applying chain rule,
(c).
Finding the exact values of k and b.
As per the above parts in (a) and (b), the initial conditions are
y'(1) = 0 and y'(2) = 3
And the equations were
Now putting the initial conditions in the equation y'(1)=0
2kbcos(b) = 0
cos b = 0 (Since, k and b cannot be zero)
And
y'(2) = 3
Answer:
+3.464; -3.464
Step-by-step explanation:
call A = y + 2x + 1 = 0 => y = (1 - 2x)
call B: 4y - 4(x^2) - 12x = -7
=> replace y from A to B =>
- 4(1 - 2x) - 4(x^2) - 12x = -7
- 4 - 8x - 4(x ^ 2) - 12x = -7
- -8x - 4(x ^ 2) - 12x = -7 - 4 = -11
- -4(x^2) - (8x - 12x) = -11
- -4(x^2) + 4x = -11
- -4(x^2) + 4x + 11 = 0
=> get delta Δ = (-4^2) - 4*(-4 * 11) = 192
=> Δ > 0 => got 2 No
=> x1 = =
= -1.232
=> x2 = =
= 2.232
=> replace x from B into A
=> y1 = (1 - 2x) = (1 - 2 * -1.232) = 3.464
=> y2 = (1 - 2x) = (1 - 2 * 2.232) = - 3.464