A wheel has a radius of 15 cm . approximately how far does it travel in 4 revolutions?

Find BC.( the value of x) 1. 2.3 cm 2. 2 cm 3. 1.5 cm 4. 1.2 cm PLEASE HELP!!

It's a proportion.
2 is to x as 15 is to 9, or 2/x = 15/9.
Multiply the means and extremes
15x = 18
divide both sides by 15 to get x by itself.
x = 1.2

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3 years ago

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Y = f(x) has the derivative f'(x) = (x + 1)²(x + 3)(x² + 2mx + 5) with ∀x∈i.

maksim [4K]

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Answer:

m ≥ -√5

Step-by-step explanation:

If g(x) = f(|x|) for x∈i, then g(x) = f(x) for x∈ℝ: x ≥ 0.

f'(x) is 5th-degree, so f(x) is 6th-degree, meaning it is generally U-shaped. Since we're only concerned with x ≥ 0, we want to make sure f'(x) has no real zeros of odd multiplicity such that x > 0. The given factors of f'(x) make it have real zeros at x = -3 and x = -1.

For the last factor, (x² +2mx +5) to have no positive real zeros of odd multiplicity, we must have m ≥ 0 or the discriminant ≤ 0. The discriminant is ...

d = b² -4ac = (2m)² -4(1)(5) = 4m² -20 . . . . . discriminant of the last factor

d ≤ 0 . . . . . . . . . . the condition for no real zeros

4m² -20 ≤ 0

m² -5 ≤ 0 . . . . . . divide by 4

m² ≤ 5 . . . . . . . . .add 5

|m| ≤ √5 . . . . . . . take the square root

This tells us there will be a positive real zero of multiplicity 2 in f'(x) when m = -√5, and there will be no positive real zeros for -√5 < m < 0

There will be no odd-multiplicity positive real zeros in the derivative function f'(x) as long as m ≥ -√5. This means the slope of f(x) is non-negative for x ≥ 0, hence f(|x|) has its only minimum at x=0.

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Additional comment

The multiplicity of the zeros of f'(x) is important because the derivative will only change sign where the multiplicity is odd. When the discriminant of (x²+2mx+5) is zero, the associated positive real zero will have multiplicity 2, hence f'(x) will not change sign there.