Find the area of the figure

Paul is three times as old as Mary and Peter is four times as old as Mary,If the sum of their ages is 64, how old are Peter,Paul

uysha [10]

Let Mary's age be x.

Paul is three times as old as Mary, so Paul's age would be 3x.

Peter is four times as old as Mary, so Peter's age would be 4x.

The sum of their ages is 64
⇒ Paul's age+ Peter's age+ Mary's age= 64
⇒ 3x+ 4x+ x= 64
⇒ 8x= 64
⇒ x= 64/8
⇒ x= 8

Paul's age is: 3x= 3*8= 24

Peter's age is: 4x= 4*8= 32

Final answer: Paul is 24 years old.
Peter is 32 years old.
Mary is 8 years old.

Hope this helps~

7 0

3 years ago

A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner

frozen [14]

Answer:

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Step-by-step explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner $I_{corner}$ is:

$I_{corner} = \int\limits \int\limits_R (x^2+y^2) \rho d A \\ \\ I_{corner} = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx$

where :

(a and b are the length and the breath of the rectangle respectively )

$I_{corner} = \rho \int\limits^a_0 {x^2y}+ \frac{y^3}{3} |^ {^ b}_{_0} \, dx$

$I_{corner} = \rho \int\limits^a_0 (bx^2 + \frac{b^3}{3})dx$

$I_{corner} = \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}$

$I_{corner} = \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]$

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Thus; the moment of inertia of the lamina about one corner is $I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

7 0

3 years ago

Graph each piecewise function. Then identify the properties.

Alexxx [7]

Answer:

Domain: (-∞, -2) U [-2, 1] U (1, ∞)

Range: (-∞, 2) U [-3, -∞)

Step-by-step explanation:

This function is composed of three polynomial functions, a positive slope line $m = \frac{1}{2}$ , whose domain is all real numbers, a horizontal line, and a negative slope line $m = -1$ whose domain is all real numbers.