Factor 4m - 10m^3<br> PLZ HELP

Which of the following describes the graph of a linear function?

Musya8 [376]

Its y-values increase at a nonconstant rate as its x-value increases

4 0

3 years ago

Read 2 more answers

(a) Find the unit tangent and unit normal vectors T(t) and N(t).

andrey2020 [161]

Answer:

a. i. (i + tj + 2tk)/√(1 + 5t²)

ii. (-5ti + j + 2k)/√[25t² + 5]

b. √5/[√(1 + 5t²)]³

Step-by-step explanation:

a. The unit tangent

The unit tangent T(t) = r'(t)/|r'(t)| where |r'(t)| = magnitude of r'(t)

r(t) = (t, t²/2, t²)

r'(t) = dr(t)/dt = d(t, t²/2, t²)/dt = (1, t, 2t)

|r'(t)| = √[1² + t² + (2t)²] = √[1² + t² + 4t²] = √(1 + 5t²)

So, T(t) = r'(t)/|r'(t)| = (1, t, 2t)/√(1 + 5t²) = (i + tj + 2tk)/√(1 + 5t²)

ii. The unit normal

The unit normal N(t) = T'(t)/|T'(t)|

T'(t) = dT(t)/dt = d[ (i + tj + 2tk)/√(1 + 5t²)]/dt

= -5ti/√(1 + 5t²)⁻³ + [-5t²j/√(1 + 5t²)⁻³] + [-10tk/√(1 + 5t²)⁻³]

= -5ti/√(1 + 5t²)⁻³ + [-5t²j/√(1 + 5t²)⁻³] + j/√(1 + 5t²)+ [-10t²k/√(1 + 5t²)⁻³] + 2k/√(1 + 5t²)

= -5ti/√(1 + 5t²)⁻³ - 5t²j/[√(1 + 5t²)]⁻³ + j/√(1 + 5t²) - 10t²k/[√(1 + 5t²)]⁻³ + 2k/√(1 + 5t²)

= -5ti/√(1 + 5t²)⁻³ - 5t²j/[√(1 + 5t²)]⁻³ - 10t²k/[√(1 + 5t²)]⁻³ + j/√(1 + 5t²) + 2k/√(1 + 5t²)

= -(i + tj + 2tk)5t/[√(1 + 5t²)]⁻³ + (j + 2k)/√(1 + 5t²)

We multiply by the L.C.M [√(1 + 5t²)]³ to simplify it further

= [√(1 + 5t²)]³ × -(i + tj + 2tk)5t/[√(1 + 5t²)]⁻³ + [√(1 + 5t²)]³ × (j + 2k)/√(1 + 5t²)

= -(i + tj + 2tk)5t + (j + 2k)(1 + 5t²)

= -5ti - 5²tj - 10t²k + j + 5t²j + 2k + 10t²k

= -5ti + j + 2k

So, the magnitude of T'(t) = |T'(t)| = √[(-5t)² + 1² + 2²] = √[25t² + 1 + 4] = √[25t² + 5]

So, the normal vector N(t) = T'(t)/|T'(t)| = (-5ti + j + 2k)/√[25t² + 5]

(b) Use Formula 9 to find the curvature.

The curvature κ = |r'(t) × r"(t)|/|r'(t)|³

since r'(t) = (1, t, 2t), r"(t) = dr'/dt = d(1, t, 2t)/dt = (0, 1, 2)

r'(t) = i + tj + 2tk and r"(t) = j + 2k

r'(t) × r"(t) = (i + tj + 2tk) × (j + 2k)

= i × j + i × 2k + tj × j + tj × 2k + 2tk × j + 2tk × k

= k - 2j + 0 + 2ti - 2ti + 0

= -2j + k

So magnitude r'(t) × r"(t) = |r'(t) × r"(t)| = √[(-2)² + 1²] = √(4 + 1) = √5

magnitude of r'(t) = |r'(t)| = √(1 + 5t²)

|r'(t)|³ = [√(1 + 5t²)]³

κ = |r'(t) × r"(t)|/|r'(t)|³ = √5/[√(1 + 5t²)]³

8 0

2 years ago

7/4x^2 -2 = - 0.05x + 4

nata0808 [166]

Answer:

$x1=\frac{-2-2\sqrt{29401} }{245} x2=\frac{-2+2\sqrt{29401} }{245}$

Step-by-step explanation:

Step One: Convert

49/16x^2-2=-0.05x+4

Step Two: Multiply Both Sides by 80

245x^2-160=-4x+320

Step Three: Move everything to the left

245x^2+4x-480=0

Final Step: Quadratic Formula

$x1=\frac{-2-2\sqrt{29401} }{245} x2=\frac{-2+2\sqrt{29401} }{245}$

6 0

3 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

3 years ago

Unit of Measurement, a punk/disco band, was signed up to play a benefit

vova2212 [387]

D because they played for two days so that makes you multiply the number of seconds by the same number

7 0

3 years ago

Factor 4m - 10m^3 PLZ HELP