Using algebra, find the point at which the line k(x) = 5x - 1 intersects with the line h(x) = -3x - 1 ?

Mathematics

1 answer:

vichka [17]1 year ago

8 0

The point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)

Given: k(x) = 5x - 1, h(x) = -3x - 1

We need to find the point(if any) at which these two lines k and h meets.

To find point of intersection(if any), we need to set the functions equal as at the point of intersection the (x, y) value will be same for both of the lines.

Therefore, k(x) = h(x)

=> 5x - 1 = -3x - 1

=> 8x = 0

=> x = 0

k(x=0) = 5 * 0 - 1 = -1

Hence the point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)

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3 0

1 year ago

If θ=0rad at t=0s, what is the blade's angular position at t=20s

babunello [35]

The attached figure represents the relation between ω (rpm) and t (seconds)
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
              ω = 250 (rpm) = 250 * (2π/60) = (25/3)π    rad/s
              ω = 100 (rpm) = 100 * (2π/60) = (10/3)π    rad/s

and from the figure it is clear that the operation is at constant speed but with variable levels
            ⇒   ω = dθ/dt   ⇒   dθ = ω dt

            ∴    θ = ∫₀²⁰ ω dt

while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
       ω = 0                                          from t = 0 to t = 5
       ω = 250 (rpm) = (25/3)π            from t = 5   to t = 15
       ω = 100 (rpm) = (10/3)π            from t = 15 to t = 20

∴ θ = ∫₀⁵ (0) dt   + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt

the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π

∴ θ = 0 + (250/3)π + (50/3)π = 100 π

while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad

Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50