A theater sells out all 1200 seats for a performance at $13 per ticket. the theater's total revenue that night is

Helppppppppppppp. a and b partttt

PilotLPTM [1.2K]

Answer:

a) 220.17 cm

b) 2127.12 cm²

Step-by-step explanation:

perimeter = circumference

it is the sum of the inner and the outer arc and the 2 side lines of 25.

we know a full circle has 360 degrees. but we only look at a segment of a circle of 150 out of the total of 360 degrees.

the circumference of a circle is

C = 2×pi×r

the 150 degree part is then

C = 2×pi×r×150/360 = 2×pi×r×15/36 = pi×r×15/18 =

= pi×r×5/6

the radius of the inner circle is 20.

=>

Ci = pi×20×5/6 = pi×50/3

the radius of the outer circle is 20 + 25 = 45

Co = pi×45×5/6 = pi×75/2

so, the total perimeter/ circumference is

C = Ci + Co + 25 + 25 = pi×50/3 + pi×75/2 + 50

= pi×100/6 + pi×225/6 + 50 = pi×325/6 + 50 =

= 220.17 cm

the area of the shaded shape is the area of the large circle segment minus the area of the small circle segment.

the area of a circle is

A = pi×r²

the 150 degree part is then

A = pi×r²×150/360 = pi×r²×5/12

so, for the inner circle that is

Ai = pi×20²×5/12 = pi×400×5/12 = pi×100×5/3 = pi×500/3

for the outer circle that is

Ao = pi×45²×5/12 = pi×2025×5/12 = pi×10125/12 =

= pi×3375/4

so, the total area of the shaded shape is

A = Ao - Ai = pi×3375/4 - pi×500/3 = pi×(10125-2000)/12 =

= pi×8125/12 = 2127.12 cm²

4 0

3 years ago

Para armar un escritorio, Eduardo tendrá que usar 20 tarugos, los que debe cubrir completamente con una capa de pegamento. Él ca

Degger [83]

El área superficial total de los veinte tarugos es mayor que el área superficial cubierta por el pegamento.

<h3>Como determinar si la superficie cubierta por el pegamento es suficiente</h3>

En esta pregunta debemos calcular el área superficial total de los tarugos ( $A_{s}$ ), en milímetros cuadrados, y comparar el resultado con la cifra afirmada por Eduardo. El área superficial total es:

$A_{s} = 20\cdot \pi\cdot (5\,mm)^{2}\cdot (80\,mm)$

$A_{s} \approx 125663.706\,mm^{2 }$

En consecuencia, el área superficial total de los veinte tarugos es mayor que el área superficial cubierta por el pegamento. Francisca tiene la razón. $\blacksquare$