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I like Hectictar's method better   but here is a different way.

Divide the pentagon inot 3 triangles. Find the lengths of all the sides of all the triangles.

Then find the areas, one at a time, using Heron's formula.

Sorry guys, I had failed to seet he word 'perpendicular' in the question.

I better get some new specs.

Thanks Heureka :)

Also answered here!

A right circular cone is inscribed in a right circular cylinder. The volume of the cylinder is  \(72\pi\)  cubic centimeters. What is the number of cubic centimeters in the space inside the cylinder but outside the cone? Express your answer in terms of .

The volume of the cone is 1/3 the volume of the cylinder

so the volume of the cylinder not including the cone is the other 2/3

2/3*72 = 48

the desired volume is \(48\pi\)    cubic cntimetres

Here's one way.....

We can break the shape up into right triangles and rectangles, then get the area of each piece.

Remember   the area of a triangle  =   (1/2)(length of base)(height)

Add the areas together...

6 + 5 + 2 + 9 + 5 + 20   =   47    sq units

Let the height of the cylinder be  H

Let the area of the base of the cylinder be  B

volume of cylinder  =  BH  =  72π

volume of cone  =  (1/3)BH   =   (1/3)(72π)  =  24π

volume of cylinder - volume of cone  =  72π - 24π  =  48π     (cubic centimeters)

help please

No, that's not correct! Anyone?

This same question was answered not too long ago...

Since  QR  =  PR  ,   △PQR  is isoscelese and its base angles are congruent. So...

∠PQR   =   ∠RPQ   =   48°

And vertical angles are congruent, so...

∠RPQ   =   ∠MPN   =   48°

Since  MP  =  NP  ,   △MNP  is isoscelese and its base angles are congruent. So...

∠PNM   =   ∠PMN

Since there are 180°  in every triangle...

∠MPN + ∠PMN + ∠PNM  =  180°

                                                         Plug in  48°  for  ∠MPN  and plug in  ∠PMN  for  ∠PNM

48° + ∠PMN + ∠PMN  =  180°

                                                         Combine like terms.

48° + 2(∠PMN)  =  180°

                                                         Subtract  48°  from both sides.

2(∠PMN)  =  132°

                                                         Divide both sides by  2 .

∠PMN  =  66°

A polynomial equation may have as many solutions as its degree, but sometimes some of the solutions are the same. We can find how many solutions each equation has by finding those solutions.

First one:

(x + 8)(x + 8)  =  0        Set each factor equal to zero.

x + 8  =  0      or      x + 8  =  0

x  = -8            or      x  =  -8

Notice that these two solutions are exactly the same, so really the equation just has one solution:

x  =  -8

Here is another way to work it:

(x + 8)(x + 8)  =  0

(x + 8)²  =  0               Take the ± square root of both sides.

x + 8  =  ±√0

x + 8  =  ±0                  Zero is neither positive nor negative, so we can drop the  ± .

x + 8  =  0

x  =  -8

Second one:

x² + 12x + 40  =  4     Subtract  4  from both sides of the equation.

x² + 12x + 36  =  0     Now factor the left side of the equation...

(x + 6)(x + 6)  =  0      This is just like the last problem now. Set each factor equal to zero.

x + 6  =  0     or     x + 6  =  0

x  =  -6          or     x  =  -6

These two solutions are identical, so really there is just one solution. It is:

x  =  -6

Is the answer 30.98 ??

A transformation in which a geometric figure is reflected across a line, creating a mirror image. That line is called the axis of reflection.

a function that moves an object a certain distance- a function that moves an object a certain distance.

1. We can calculate the mean by adding all the numbers and dividing it by how many numbers there are.

So, 5+8+9+11+7+7+8+9+12+7=83/10=8.3 , thus \(\boxed{8.3}\)

This is a case of the flagpole problem which can be solved with the formula \(ab/(a + b)\)where a and b are the lengths of the sides

Therefore,  the length of \(EF\)  is \(60\)  

another way to do this w/o the formula is graphing it

Great solution, both hectictar and CPhill!

\(f(x)=x+2\)          and          \(f^{-1}(x)=x-2\)

\(g(x)=x/3\)             and          \(g^{-1}(x)=3x\)

There may a better way to do this, but here is what I got:

\(\begin{array}\ &&f(g^{-1}(f^{-1}(f^{-1}(g( {\color{Red}f(19)} ))))) \\ =&&f(g^{-1}(f^{-1}(f^{-1}(g( {\color{Red}19+2} )))))\\ =&&f(g^{-1}(f^{-1}(f^{-1}(g( {\color{Red}21} )))))\\ =&&f(g^{-1}(f^{-1}(f^{-1}( {\color{Orange}g( 21 )} ))))\\ =&&f(g^{-1}(f^{-1}(f^{-1}( {\color{Orange}21/3} ))))\\ =&&f(g^{-1}(f^{-1}(f^{-1}( {\color{Orange}7} ))))\\ =&&f(g^{-1}(f^{-1}( {\color{orange}f^{-1}(7)} )))\\ =&&f(g^{-1}(f^{-1}( {\color{orange}7-2} )))\\ =&&f(g^{-1}(f^{-1}( {\color{orange}5} )))\\ =&&f(g^{-1}( {\color{Green}f^{-1}( 5 )} ))\\ =&&f(g^{-1}( {\color{Green}5-2} ))\\ =&&f(g^{-1}( {\color{Green}3} ))\\ =&&f({\color{Blue}g^{-1}( 3 )})\\ =&&f({\color{Blue}3(3)})\\ =&&f({\color{Blue}9})\\ =&&{\color{Fuchsia}f(9)}\\ =&&{\color{Fuchsia}9+2}\\ =&&{\color{Fuchsia}11} \end{array}\)

Here is another way to think about it  

If we let the side facing us be the base, then we look at the shape like a prism....

surface area  =  (perimeter of base)(height)   +  2(area of base)

perimeter of base  =  7 + 12 + 7 + 5 + 13   =   44    ft

height  =  16 ft

area of base  =  area of triangle + area of rectangle

area of base  =  (1/2)(12)(5) + (12)(7)

area of base  =  30 + 84

area of base  =  114          ft²

surface area  =  ( 44 ft )( 16 ft )  +  2( 114 ft² )

surface area  =  704  ft²  +  228  ft²

surface area  =  932  ft²

A little tricky  !!!

I  to D   we go back 5

5 to 7   we go up 2

So   (x - 5, x + 2)  ⇒ "b"

10 left, 3 down   =  (x - 10, y - 3)  ⇒  "d"

We only need to look at one vertex and its after-image....look at C, C'

C =   (2, - 2)      C'  =  ( -2, -7)

So    (  x - 4,  y  - 5)  ⇒  "b"

lateral area of cone   =   pi * slant height * radius of the base

The radius of the base is  3  inches.

It looks like  6 inches  is the slant height.

lateral area of cone   =   pi * 6 * 3   in²

lateral area of cone  =   18pi   in²

lateral area of cylinder   =   circumference of base * height

lateral area of cylinder   =   2pi * radius of base * height

The radius of the base is  3 inches.

The height is  5 inches.

lateral area of cylinder   =   2pi * 3 * 5  in²

lateral area of cylinder   =   30pi  in²

area of base   =   pi * radius²

area of base   =   pi * 3²   in²

area of base   =   9pi  in²

total surface area   =   lateral area of cone + lateral area of cylinder + area of base

total surface area   =   18pi  in²   +   30pi  in²   +   9pi  in²

total surface area   =   57pi  in²

total surface area   ≈   57(3.14)  in²

total surface area   ≈   178.98  in²

Top  13 * 16  =  208

Two triangles at front and back at the top  =  5 * 12   =  60

" Left Side"  at the top  = 5 * 16  =  80

Two sides  of 16*7  =  2*16*7  =  224

Front and back at the bottom part  =  2*12*7  = 168

Bottom  = 12 * 16  =  192

Adding all of these

208 + 60 + 80 + 224 + 168 + 192  =

932  ft^2

oh whoops. lemmie fix that.

How are we supposed to use pi in the problem. And what do we do?

-5  + something  =  0   ⇒  something  = 5

-5  +  something  =   2 ⇒  something  = 7

So     (x + 5 , y + 7)

U'  = (-5 +  5, 1 + 7)  =   (0, 8)   =  "d"

3 right, 4 down ⇒  "a"

(x - 6, y + 4)   ⇒  "d"

YIPPPEEEEE!!!!