If a^20=(a^n)^m what are the values for m and n

A cylinder shaped can needs to be constructed to hold 500 cubic centimeters of soup. The material for the sides of the can costs

iogann1982 [59]

Answer:

r=3.628cm

h=12.093cm

Step-by-step explanation:

For this problem we are going to use principles, concepts and calculations from multivariable calculus; mainly we are going to use the Lagrange multipliers method. This method is thought to help us to find a extreme value of a multivariable function 'F' given a restriction 'G'. F represents the function that we want to optimize and G is just a relation between the variables of which F depends. The Lagrange method for just one restriction is:

$\nabla F=\lambda \nabla G$

First, let's build the function that we want to optimize, that is the cost. The cost is a function that must sum the cost of the sides material and the cost of the top and bottom material. The cost of the sides material is the unitary cost (0.03) multiplied by the sides area, which is $A_s=2\pi rh$ for a cylinder; while the cost of the top and bottom material is the unitary cost (0.05) multiplied by the area of this faces, which is $A_{TyB}=2\pi r^2$ for a cylinder.

So, the cost function 'C' is:

$C=2\pi rh*0.03+2\pi r^2*0.05\\C=0.06\pi rh+0.1\pi r^2$

The restriction is the volume, which has to be of 500 cubic centimeters:

$V=500=\pi r^2h\\500=\pi hr^2$

So, let's apply the Lagrange multiplier method:

$\nabla C=\lambda \nabla V\\\frac{\partial C}{\partial r}=0.06\pi h+0.2\pi r\\\frac{\partial C}{\partial h}=0.06\pi r\\\frac{\partial V}{\partial r}=2\pi rh\\\frac{\partial V}{\partial h}=\pi r^2\\(0.06\pi h+0.2\pi r,0.06\pi r)=\lambda (2\pi rh,\pi r^2)$

At this point we have a three variable (h,r, λ)-three equation system, which solution will be the optimum point for the cost (the minimum). Let's write the system:

$0.06\pi h+0.2\pi r=2\lambda \pi rh\\0.06\pi r=\lambda \pi r^2\\500=\pi hr^2$

(In this kind of problems always the additional equation is the restricion, in this case, V=500).

Let's divide the first and second equations by π:

$0.06h+0.2r=2\lambda rh\\0.06r=\lambda r^2\\500=\pi hr^2$

Isolate λ from the second equation:

$\lambda =\frac{0.06}{r}$

Isolate h from the third equation:

$h=\frac{500}{\pi r^2}$

And then, replace λ and h in the first equation:

$0.06*\frac{500}{\pi r^2} +0.2r=2*(\frac{0.06}{r})r\frac{500}{\pi r^2} \\\frac{30}{\pi r^2}+0.2r= \frac{60}{\pi r^2}$

Multiply all the resultant equation by $\pi r^{2}$ :

$30+0.2\pi r^3=60\\0.2\pi r^3=30\\r^3=\frac{30}{0.2\pi } =\frac{150}{\pi}\\r=\sqrt[3]{\frac{150}{\pi}}\approx 3.628cm$

Then, find h by the equation $h=\frac{500}{\pi r^2}$ founded above:

$h=\frac{500}{\pi r^2}\\h=\frac{500}{\pi (3.628)^2}=12.093cm$

4 0

3 years ago

How do you convert fractions to decimals?

blsea [12.9K]

Answer:

divide the fractions

Step-by-step explanation:

lets say we have 23/5

we can divide 23 by 5 and get a decimal of 4.6

or lets say we got 1/5

divide 1 by 5 and you will get 0.2

4 0

3 years ago

Please help me do my homework I have been stuck on this for ages!!

jek_recluse [69]

10%. The thing is telling me to use twenty characters ignore this lol

8 0

3 years ago

Can someone help me with number 3 please??!!

ipn [44]

There are 2 way to solve this.

one using Pythagoras theorem and 2nd using trigonometry

so lets solve it by both

using Pythagoras theorem we know

base^2 + perpendicular^2 = hypotanes^2

6^2 + x^2 = 12^2

36 + x^2 = 144

x^2 = 144- 36 = 108

x = √(108) = √( 2×2×3×3×3)

= (2×3) √ (3) = 6 √3

so answer is option 2

bow lets use trigonometry

we know
sin theta = perpendicular / hypotanes
sin 60 = x /12
x = 12 × sin 60
we kNow sin 60 = √3/ 2
so
x = 12×√3 /2 = 6√3

7 0

3 years ago

How much fabric is needed to make this

abruzzese [7]

Answer:

f

Step-by-step explanation:

6 0

3 years ago