What is the volume of the cube below?

I need the work and answer. Please and thank you

AlladinOne [14]

Answer:

11.1

Step-by-step explanation:

mark brainliest

7 0

2 years ago

A gumball machine has 150 red gumballs. If the red gumballs are 25% of the total number of gumballs, how many gumballs are in th

WINSTONCH [101]

Answer:

Step-by-step explanation:

Because 25% is the same as 1/4. We can do the same thing as if we were changing 1/4 into a whole 1/4·4=1. So 25%·4=1. So 150·4 is our answer. So 150·4=600. There are 600 gumballs in the machine. :D

7 0

2 years ago

Read 2 more answers

5. ¿A qué fracción es equivalente el número 27

nika2105 [10]

Answer:

Step-by-step explanation:

8 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. $\blacksquare$

<h3>Observación</h3>

El enunciado omite las dimensiones <em>geométricas</em> de cada tarugo (5 milímetros de radio y 80 milímetros de longitud).

Para aprender más sobre áreas, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/391394

4 0

2 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago