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3 years ago

How do i evaluate 16^0.25 ?

Mathematics

1 answer:

KengaRu [80]3 years ago

3 0

16^0.25 = 16^(1/4) =( ⁴√16)¹ = 2

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3 years ago

If a simple harmonic motion is described by x = 3 cos(πt), starting from t = 0 s, what is the first occurrence of a value of t f

poizon [28]

Answer:

The first occurence of t for which x = 0 is t = 0.5.

Step-by-step explanation:

The harmonic motion is described by the following equation.

$x(t) = 3\cos{\pi t}$

What is the first occurrence of a value of t for which x = 0?

This is t when $x(t) = 0$ . So

$x(t) = 3\cos{\pi t}$

$3\cos{\pi t} = 0$

$\cos{\pi t} = \frac{0}{3}$

$\cos{\pi t} = 0$

The inverse of the cosine is the arcosine. So we apply the arcosine function to both sides of the equality.

$\arccos{\cos{\pi t}} = \arccos{0}$

$\pi t = \frac{\pi}{2}$

$t = \frac{\pi}{2 \pi}$

$t = \frac{1}{2}$

$t = 0.5$

The first occurence of t for which x = 0 is t = 0.5.

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2 years ago

What are the approximate values of the non-integral roots of the polynomial equation?

irga5000 [103]

Answer:

-4, -2, -3-2i, -3+2i

Step-by-step explanation:

The of the polynomial equation are those values of x, for which f(x)=0.

Consider equation

$f(x)=0\\ \\(x^2+6x+8)(x^2+6x+13)=0$

By zero product property,

$x^2+6x+8=0\ \text{or}\ x^2+6x+13=0$

Solve each equation:

1. $x^2+6x+8=0$

$D=6^2-4\cdot 8=36-32=4\\ \\x_{1,2}=\dfrac{-6\pm \sqrt{4}}{2}=\dfrac{-6\pm 2}{2}=-4,\ -2$

2. $x^2+6x+13=0$

$D=6^2-4\cdot 13=36-52=-16=16i^2\\ \\x_{1,2}=\dfrac{-6\pm \sqrt{16i^2}}{2}=\dfrac{-6\pm 4i}{2}=-3-2i,\ -3+2i$

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3 years ago

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