All categories

2 years ago

What is the surface area of the cylinder with height 2 yd and radius 3 yd? Round your answer to the nearest thousandth.

Mathematics

1 answer:

olasank [31]2 years ago

3 0

<h2>Given :</h2>

height (h) = 2 yd

radius (r) = 3 yd

<h2>Solution :</h2>

$\boxed{ \mathrm{Surface \: Area = 2\pi r(h +r )}}$

$2 \times 3.14 \times 3 \times (3 + 2)$

$18.84 \times 5$

$\boxed{\mathrm{94.2 \: yd {}^{2} }}$

_____________________________

$\mathrm{ \#TeeNForeveR}$

You might be interested in

Can someone plz help me I don’t understand how to do this and I am really struggling with the online stuff

sweet [91]

Answer:

Step-by-step explanation:

x=6.28111...

100 x=628.111...

1000x=6281.111...

subtract

900x=5653

x=5653/900

density of iodine=785×5653/900≈4930.672222...

Here 2 is repeating .

3 0

2 years ago

What is the value of point G on the number line?

olya-2409 [2.1K]

A) because it is 2/3 past 1

8 0

3 years ago

Read 2 more answers

Hi I need help for my math class it’s due Saturday!

vichka [17]

Answer:

a) 21m - 0.87 = m

b) m = 0.0435 mg

21(0.0435) - 0.87 = 0.0435

0.9135 - 0.87 = 0.0435

0.0435 = 0.0435

True statement, so correct

c) 21m - 0.87 = m

20m = 0.87

m = 0.87/20

m = 0.0435 mg

4 0

3 years ago

Read 2 more answers

Ab+c=d solve for b.

Dennis_Churaev [7]

Answer:

give branliest pls

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

A company wishes to manufacture some boxes out of card. The boxes will have 6 sides (i.e. they covered at the top). They wish th

Serhud [2]

Answer:

The dimensions are, base $b=\sqrt[3]{200}$ , depth $d=\sqrt[3]{200}$ and height $h=\sqrt[3]{200}$ .

Step-by-step explanation:

First we have to understand the problem, we have a box of unknown dimensions (base $b$ , depth $d$ and height $h$ ), and we want to optimize the used material in the box. We know the volume $V$ we want, how we want to optimize the card used in the box we need to minimize the Area $A$ of the box.

The equations are then, for Volume

$V=200cm^3 = b.h.d$

For Area

$A=2.b.h+2.d.h+2.b.d$

From the Volume equation we clear the variable $b$ to get,

$b=\frac{200}{d.h}$

And we replace this value into the Area equation to get,

$A=2.(\frac{200}{d.h} ).h+2.d.h+2.(\frac{200}{d.h} ).d$

$A=2.(\frac{200}{d} )+2.d.h+2.(\frac{200}{h} )$

So, we have our function $f(x,y)=A(d,h)$ , which we have to minimize. We apply the first partial derivative and equalize to zero to know the optimum point of the function, getting

$\frac{\partial A}{\partial d} =-\frac{400}{d^2}+2h=0$

$\frac{\partial A}{\partial h} =-\frac{400}{h^2}+2d=0$

After solving the system of equations, we get that the optimum point value is $d=\sqrt[3]{200}$ and $h=\sqrt[3]{200}$ , replacing this values into the equation of variable $b$ we get $b=\sqrt[3]{200}$ .

Now, we have to check with the hessian matrix if the value is a minimum,

The hessian matrix is defined as,

$H=\left[\begin{array}{ccc}\frac{\partial^2 A}{\partial d^2} &\frac{\partial^2 A}{\partial d \partial h}\\\frac{\partial^2 A}{\partial h \partial d}&\frac{\partial^2 A}{\partial p^2}\end{array}\right]$

we know that,

$\frac{\partial^2 A}{\partial d^2}=\frac{\partial}{\partial d}(-\frac{400}{d^2}+2h )=\frac{800}{d^3}$

$\frac{\partial^2 A}{\partial h^2}=\frac{\partial}{\partial h}(-\frac{400}{h^2}+2d )=\frac{800}{h^3}$

$\frac{\partial^2 A}{\partial d \partial h}=\frac{\partial^2 A}{\partial h \partial d}=\frac{\partial}{\partial h}(-\frac{400}{d^2}+2h )=2$

Then, our matrix is

$H=\left[\begin{array}{ccc}4&2\\2&4\end{array}\right]$

Now, we found the eigenvalues of the matrix as follow

$det(H-\lambda I)=det(\left[\begin{array}{ccc}4-\lambda&2\\2&4-\lambda\end{array}\right] )=(4-\lambda)^2-4=0$

Solving for $\lambda$ , we get that the eigenvalues are: $\lambda_1=2$ and $\lambda_2=6$ , how both are positive the Hessian matrix is positive definite which means that the function $A(d,h)$ is minimum at that point.

4 0

3 years ago