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3)What is the total surface area of this prism?

2)ABCDE is a right square pyramid with DE = 6 and AC = 3sqrt5. Find the total surface area of the pyramid. 

1)ABCDEFGH is a rectangular prism with AB = 3, BC = 4  and AE = 5. Find the area of quadrilateral ACGE.

Congrats! How much of it was because of the help you got from this forum?

See the answer and the method used here:

http://mathcentral.uregina.ca/QQ/database/QQ.09.06/richard1.html

Thanks so much, CPhill! 

You can also solve it this way:

FV = PV x [1 + x]^(t*12)

140 = 21 x [1 + x]^(27*12)

140 = 21 x [1 + x]^324            divide both sides by 21

6 2/3 = [1+ x]^324                   take the log of both sides

0.82391... = 324 x log[1 +x]    divide both sides by 324

Log[1 + x] =0.002542928.....   take the exponent of both sides

[1 + x] = 10^0.002542928...

[1 + x] = 1.005872484421...    subtract 1 from both sides

x = 1.005872484421.... - 1 x 100 x 12

x = ~7.05% compounded monthly. The APR of this investment.

1. 

a)  It makes one revolution every 3 seconds....so....it makes 20 revolutions every minute

b)  The linear velocity  is  2pi  (5) ( 120 /360)  =   (10/3)pi  inches per second

The surface area of the sphere is:     

 \(4\pi r^{ 2 }=4\pi \times 5^{ 2 }\approx 314.16\)                              

The volume of the sphere is:

\(\frac { 4 }{ 3 } \pi r^{ 3 }=\frac { 4 }{ 3 } \pi \times 5^{ 3 }\approx 523.6\)

The opposite angles of a cyclic quadrilateral (a quadrilateral whose vertices touch the circumference of a circle) always add up to 180°.

Thus:

\(x+4x-20=180\)

\(5x-20=180\)

\(5x=180+20=200\)

\(x=\frac { 200 }{ 5 } =40\)

\(\angle B=(4x-20)°=(4\times 40-20)°=(160-20)°=140°\)

So, the measure of angle B is 140°.

This is an isosceles trapezoid...angle  DEB   = 45° 

We can find the height, H,  of the tapezoid as

sin(45)  =  H / 12

12 sin (45)  = H

6√2   cm = H

And the longer base length is [12 + 6√2  + 6√2]  =   [ 12  + 12√2  ] cm

So....the area  is

height  *  (sum of base lengths)  / 2 =

6√2  [  12 + (12 +12√2) ] / 2  =

6√2 [ 24  + 12√2 ] / 2  =

3√2  [ 24  + 12√2]  =

72√2  + 72  =

72 (√2 + 1)  cm^2  

Angles A  and C  are supplementary     And B and D  are supplementary [sum to 180 degrees ]

So we have

(x + 2)  + ( x - 2)  =180

2x  = 180     divide both sides by 2

x  = 90

So  A  =   90 + 2 = 92 degrees

C  =  90 - 2  =  88 degrees

D  = 90  - 10  =  80 degrees

And  B = 180  - D   =  180  - 80  = 100 degrees

In the form  y  =  mx + b.....m is the slope and b is the y intercept

The slope is negative....so....we can automaticaly eliminate A  and B

Note that the y intercept is positive

The only answer that works is  "C"

Second one

There is no horizontal asymptote here....this will only happen when the polynomials in the numerator and denominator have the same degree

LOL!!!

Here's what EP  is saying

Volume  =  (4/3)pi (radius)^3   = (4/3)pi (5)^3   =  (4/3)pi* 125 ≈ 523.6  units^3 

Surface Area =  4pi (radius)^2  =  4pi (5)^2  = 100pi  ≈  314.16 units^2

im ssorry, i dont get it TOT

We can find the height of the ttrapezoid thusly

300  =  h ( 30 + 20) /2

300 = h * 25

!2  = h

Let A  = (5,12)   B  = (25,12)  C  = (30,0)  and D  = (0,0)

The area  of triangle DBC  = (1/2) CD * 12  =  (1/2) (30) * 12  =  180  units^2     (1)

The  equation  of line segment BD  is

y  =(12/25)x

The y coordinate  of X   is the height of triangle CXD and   can be found as

y  =  (12/25) (15)   = 3/5 * 12  = 7.2

So....the area of triangle  CXD  = (1/2)(CD)(7.2)  = (1/2)(30)(7.2)  = 108 units^2     (2)

So...the area  of triangle BXC  =  (1)  - (2)  =  180  - 108  =  72 units^2

Here's a pic :

Well .....    MINUS  25.8069 is the root of at least SOME of it !

Surface area of a sphere = 4 pi r^2

Volume of a sphere = 4/3  pi r^3               r = 5 in in this case......I'll let you plug and chug

Just 'cos' they wanted to?

OPPOSITE angles of a cyclic quadrilateral ( a quadrilateral with all vertices on the circumference of a given circle) add to 180 degrees.....s o

x + (4x-20) = 180

5x=200

x = 40

B = 4x-20

       4(40) - 20 = 140 degrees