A quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic equation is y = a $x^{2}$ + bx + c, where a, b, and c are numbers and a cannot be 0

P(A) = 0.006 $a^{2}$ - 0.02a + 120

123 = 0.006- 0.02a + 120

0=0.006 $a^{2}$ - 0.02a - 3

you can use the quadratic equation formula to solve for the man's age.

A = (-b ± ( $\sqrt{b^{2} - 4*a*c}$ ) ) / (2a)

A = (0.02 ± $\sqrt{(-0.02)^{2} - 4*0.006*(-3)}$ / (2*0.006)

A = (0.02 ± $\sqrt{0.0076}$ ) / 0.012

A = 9 , -5.67

Age of the man will be 9 years

To learn more about quadratic equation here

brainly.com/question/17177510?referrer=searchResults

#SPJ4

5 0

2 years ago

Fluid ounces measure _____?<br> Volume<br> Weight<br> Viscosity

ladessa [460]

Fluid ounces measure in volume

5 0

4 years ago

Read 2 more answers

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.55 times a second. A tack is stuck in the tire a

earnstyle [38]

Answer:

Tangential speed = 5.72 m/s

Centripetal acceleration = $91.6\text{ m/s}{}^2$

Explanation:

The tangential speed, V, is given by

$v=\omega r$

where $\omega$ is the angular speed and is given by $2\pi f$ (f is the angular frequency or frequency of rotation)

Thus,

$v=2\pi f r = 2\times3.14\times2.55\times0.357 = 5.72\text{ m/s}$

The centripetal acceleration,a, is given by

$a=\dfrac{v^2}{r}$

$a=\dfrac{5.72^2}{0.357} = 91.6\text{ m/s}{}^2$

7 0

4 years ago

Read 2 more answers

An airplane went from 120 m/s to 180 m/s in 4.0 seconds. What was its acceleration?

scoundrel [369]

<span>15 m/s^2 The first thing to calculate is the difference between the final and initial velocities. So 180 m/s - 120 m/s = 60 m/s So the plane changed velocity by a total of 60 m/s. Now divide that change in velocity by the amount of time taken to cause that change in velocity, giving 60 m/s / 4.0 s = 15.0 m/s^2 Since you only have 2 significaant figures, round the result to 2 significant figures giving 15 m/s^2</span>

3 0

3 years ago