15x²+16x+4 =0 (ax² +bx +c=0)

Δ = b²-4ac =256 - 4×15×4 =16

x1 = (-b+√Δ) / 2a = (-16+√16) / 30 =( -16+4) / 30 = -12/30 = - 2/5
x2 = (-b -√Δ) / 2a = (-16 -√16) / 30 = (-16 -4) /30 = -20/30 = -2/3

4 0

3 years ago

Read 2 more answers

A guitar string is plucked at a distance of 0.6 centimeters above its resting position and then released, causing vibration. The

liubo4ka [24]

Answer:

The equation that represents the motion of the string is given by:

$y =Ae^{-kt}\cos(2\pi ft)$ .....[1] where t represents the time in second.

Given that: A = 0.6 cm (distance above its resting position) , k = 1.8(damping constant) and frequency(f) = 105 cycles per second.

Substitute the given values in [1] we get;

$y =0.6e^{-1.8t}\cos(2\pi 105t)$

$y =0.6e^{-1.8t}\cos(210\pi t)$

(a)

The trigonometric function that models the motion of the string is given by:

$y =0.6e^{-1.8t}\cos(210\pi t)$

(b)

Determine the amount of time t that it takes the string to be damped so that $-0.24\leq y \leq0.24$

Using graphing calculator for the equation

$y =0.6e^{-1.8x}\cos(210\pi x)$

let x = t (time in sec)

Graph as shown below in the attachment:

we get:

the amount of time t that it takes the string to be damped so that $-0.24\leq y \leq0.24$ is, 0.5 sec

4 0

3 years ago

The mean length of 5 childrens' index finger is 5.6cm.

Brilliant_brown [7]

The mean length (rounded to 2 DP) of these 8 people's index finger is 5.83

<h3>What is the mean length (rounded to 2 DP) of these 8 people's index finger?</h3>

The given parameters are:

Children

Mean = 5.6 cm

Frequency = 5

Adult

Mean = 6.2 cm

Frequency = 3

The mean length of these 8 people's index finger is calculated as:

$\bar x = \frac{\sum fx}{\sum f}$

So, we have

Mean = (5.6 * 5 + 6.2 * 3)/(5 + 3)

Evaluate

Mean = 5.83

Hence, the mean length (rounded to 2 DP) of these 8 people's index finger is 5.83