What is the surface area of the cube

Mathematics

1 answer:

madam [21]3 years ago

3 0

There are six faces on the cube, each with a surface area of (4.5 m)^2 = 20.25 m^2. They add up to the amount given by the first selection.

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Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve

Salsk061 [2.6K]

Answer:

$\large \boxed{2\pi(15 - \ln16)}$

Step-by-step explanation:

A 90° portion of your volume is shown below.

Let's use the shell method and decompose the solid of revolution into cylindrical shells.

The volume of each cylindrical shell is

dV = 2πrh dx

where

r = ln16 - x

h = eˣ

dV = 2π(ln16 - x)eˣdx

$\begin{array}{rcl}V& =& 2\pi \int_{0}^{\ln16}(\ln16 - x)e^{x}dx\\\\&=&2\pi \int_{0}^{\ln16}(\ln 16)e^{x}dx -2\pi \int_{0}^{\ln16} xe^{x}dx\\\\&=& 2\pi \ln16 \int_{0}^{\ln16}e^{x}dx - 2\pi\int_{0}^{\ln16}xe^{x}dx\\\\\end{array}$

Let's integrate the two terms separately.

$\begin{array}{rcl}A &=& 2\pi \ln16 \int_{0}^{\ln16}e^{x}dx\\\\&=& 2\pi\ln16[e^{x}]_{0}^{\ln16}\\\\&=& 2\pi\ln16(16 - 1)\\&= &\mathbf{30\pi\ln1}\\\\B &=&2\pi\int_{0}^{\ln16}xe^{x}dx\\\end{array}$

Use the integration by parts formula

$\int u dv = uv - \int v du$

$\begin{array}{rcl}uv - \int v du &=&\\B &=& 2\pi[xe^{x}]_{0}^{\ln16} - 2\pi \int_{0}^{\ln16} e^{x}dx\\\\&=&2\pi[xe^{x}]_{0}^{\ln16}-2\pi [e^{x}]_{0 }^{\ln16}\\\\&= &2\pi(16\ln16- 0) -2\pi(16 - 1)\\&=& \mathbf{32\pi \ln16 -30\pi}\\A - B &= &30\pi\ln16 - 32\pi \ln16 +30\pi\\&=& 30\pi - 2\pi \ln16\\&=&\mathbf{2\pi(15 - \ln16)}\end{array}\\\text{The volume of the solid is $\large \boxed{\mathbf{2\pi(15 - \ln16)}}$}$

7 0

3 years ago

What is the simplified form of the expression (6^-4)^-9/6^6?

aleksley [76]

Answer:

$(\frac{1}{1296} )^{\frac{729}{64} }$

Step-by-step explanation:

I am assuming this problem reads $(6^{-4}) ^{-\frac{9}{6}^6 }$

So let's start on the outside and work our way towards the middle
Starting with the $(-\frac{9}{6} )^{6}$ , this can be simplified to $\frac{729}{64}$

Now we have $(6^{-4})^\frac{729}{64}$

Now simplifying what is inside the parenthesis we get $\frac{1}{1296}$
Combining what we have the simplified for of the problem becomes: $(\frac{1}{1296} )^{\frac{729}{64} }$