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Did you create this problem and do you know the answer? 

What is 'r'? If you give us a number that will make it easier to solve.

We need to get a common denominator. Let's multiply the first fraction by  3/3 .

$\frac{\sqrt5}{10}+\frac{\sqrt2}{6}+2 \\~\\ =\,\frac{\sqrt5}{10}\cdot\frac33+\frac{\sqrt2}{6}+2 \\~\\ =\,\frac{3\sqrt5}{30}+\frac{\sqrt2}{6}+2$

Now let's multiply the second fraction by  5/5 .

$=\,\frac{3\sqrt5}{30}+\frac{5\sqrt2}{30}+2$

And multiply the  2  by  30/30 .

$=\,\frac{3\sqrt5}{30}+\frac{5\sqrt2}{30}+\frac{60}{30}$

Now we can add the fractions together.

$=\,\frac{3\sqrt5+5\sqrt2+60}{30}$

surface area of cylinder  =  (circumference of base)(height) + 2(area of base)

surface area of cylinder  =  2 pi r h  +  2 pi r²               r  =  60/2  =  30    and    h = 4 * 60  =  240

surface area of cylinder  =  2 pi (30)(240)  +  2 pi (30)²

surface area of cylinder  =  14400pi + 1800pi   =   16200pi   ≈   50894 square mm

surface area of box  =  2LW + 2LH + 2HW

And  L = 60 + 60 = 120 mm ,  W = 60 + 60 = 120 mm ,  H = 60 mm

surface area of box  =  2(120)(120) + 2(120)(60) + 2(60)(120)

surface area of box  =  28800 + 14400 + 14400   =   57600 square mm

50894  is less than  57600 , so the cylinder has a smaller surface area.

V  =  (1/3) (17.2)^2 * 24.4   ≈  2406.17 cm^3

Thanks, sm!!!!.... we could use Heron's here, but here's a slightly easier approach...IMHO...!!!

Draw an altitude of the triangle......and the altitude of an isosceles triangle will divide the base in half

So....we will have two right triangles 

Looking at either of these.....they will have a hypotenuse of 13cm and one leg = 1/2 of the base   = 5cm

And the other leg will  be   sqrt (13^2 - 5^2)  = sqrt (144)  = 12cm  = the height (altitude) of the triangle

So....the area  is  

1 / 2  base  * height  =

5 cm  *  12 cm  =

60 cm^2

If we extended the top base of the figure by 8m and drew a perpendicular line to the base...the additional area would create a right "Pythagorean Triple" triangle with one of the legs  = 8m and a hypotenuse of 17m.....and the other leg = sqrt (17^2 - 8^2)  = sqrt (225) =  15m  = the length of the fence

Thus....disregarding the right triangle, we have a trapezoid with a height of 15m and two bases of 12m and 20m

So....the  area of a trapezoid  =

(height) / 2  *  (sum of the bases )   =

15 / 2  * ( 12 + 20)  =

7.5 *  32   =

240 m^2

So...we need three cans of weedkiller  =  3 * 19.75  =  59.25 pounds  (total cost) 

I use a^2 + b^2 = c^2  (26 - 11)^2 + 8^2 = x^2, x = 17 ab = 17

Using a^2 + b^2 = c^2 again, 17^2 + 12^2 = x^2  x=square root of 433

Because it is an iscocoles triangle, we know two sides are 13 and one side is 10. 

Using heron's fomula: http://www.mathwarehouse.com/geometry/triangles/area/herons-formula-triangle-area.php 

The answer is 60. 

e^x   has the unique property that it is its own derivative...thus.....

f(x)  =  e^x

And

f¹⁰⁰ (x)  =  e^x

Thank You

Thanks that is a lot of work!! Ur da best!!

Alrighty Thanks!

Why thank you!

The first will produce vertical lines through a and -a on the x axis

The second will produce horizontal lines through b and -b on the y axis

Very nice, Melody !!!!

Let D be 1/2 of the  Diistance

And 40 min  2/3 hr

So   using Distance / Rate  = Time in Hours .......we have that

D/3  =  D/5   + 2/3    simplify

D/3  +  [3D + 10] / 15     cross-multiply

15D  = 3[ 3D + 10]

15D  =  9D + 30     subtract 9D from both sides

6D  = 30   divide both sides by 6

D  = 5 miles

So...the total distance is twice this  =  10 miles