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t is 10. given t/5=2 multiply both sides by 5, then T = 10

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Q: What is T if T over 5 equals 2?

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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

5 = Toes on a Foot

figure it out! But you can''t figure out N in this equation

(f(g(t)))'=f'(g(t)) x g'(t) y=sqrt(3x-5) here f(t)=sqrt(t) and g(t)=3t-5, f'(t)=1/(2 sqrt(t)) and g'(t) = 3 So y'= f'(g(x)) x g'(x) = 1/(2 sqrt(3x-5) x 3 = 3/(2 sqrt(3x-5))

Related questions

(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

50

It Takes 2 to Tango

4

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

5t - 1 = -11 Therefore, 5t = -10 t = -10/5 t = -2

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