t is 10. given t/5=2 multiply both sides by 5, then T = 10
If -t = 5, then t = -5
2 (t = -1)
Depends on which question you're asking: 1) (5t-5)/-10=70 5t-5=-700 5t=-695 t=-139 2) (5(t-5))/-10=70 5(t-5)=-700 5t-25=-700 5t=-675 t=-135
t=-2
4
(t-7)/w = (-3-7)/-2 = -10/-2 = +5 ( minus over minus = plus)
If -t = 5, then t = -5
2 (t = -1)
Depends on which question you're asking: 1) (5t-5)/-10=70 5t-5=-700 5t=-695 t=-139 2) (5(t-5))/-10=70 5(t-5)=-700 5t-25=-700 5t=-675 t=-135
t=-2
y=x3+ 2x, dx/dt=5, x=2, dy/dt=? Differentiate the equation with respect to t. dy/dt=3x2*dx/dt Substitute in known values. dy/dt=3(2)2 * (5) dy/dt=60
Assuming the question refers to x(t) = 3.4*cos(5*pi*t) - pi, the periodicity of the cos function is 2*pi So 5*pi*t = 2*pi or 5*t = 2 or t = 0.4
4
50
sin(t) = 2/3 sin2(t) + cos2(t) = 1 so cos(t) = ± sqrt[1 - sin2(t)] but because t is in the first quadrant, cos(t) > 0 so cos(t) = + sqrt[1 - sin2(t)] = sqrt[1 - 4/9] = sqrt[5/9] = sqrt(5)/3 Then sec(t) = 1/cos(t) = 1/sqrt(5)/3 = 3/sqrt(5) = 3*sqrt(5)/5
It Takes 2 to Tango
5t - 1 = -11 Therefore, 5t = -10 t = -10/5 t = -2