CPhill

True area  =   pi (20/2)^2  = 100 pi cm^2

If he underestimates the circumference by 20%....the measured diameter  is 20 * .80  = 16 cm

And the  area in this case   =     pi (16/2)^2  = pi * 8^2  =  64 pi

So.....the percent error in the area in this case   =   l  100 pi - 64pi  l / l 100 pi l  = 36 / 100  = 36%

If he overestimates the circumference by 20%....the measured diameter  = 20 * 1.20  = 24 cm

And the area in this case = pi (24/2)^2  = pi * 12^2  = 144 pi

So.....the percent error in this case  =  l  144 pi - 100 pi  l  / l  100 pi  l   =   44 / 100  =  44%

( √5) ^-2    is the same as    1  / (√5)²     =   1 / [ √5 * √5 ]   =   1  / 5

23. Number of tickets.    Total Cost

                   2.                            $7.  

                   4.                            $14

                   6.                            $21

These are all pretty much the same type of linear problem .......let x be the Number of Tickets and y be the Total Cost

Note that  as y changes by $7, x changes by 2.....so...the slope of our line  = 7/2

And letting one point  be  (2, 7)....we have the following equation :

y = (7/2)(x - 2) + 7

24, 25  and 27 are similar to 23

26.   Total cost.      Change from $10

            $10.00.                    $0

            $9.00.                      $1.00 

            $7.50.                      $2.50

This one is a little tricky......let the Change from $10 be y  and the Total Cost  = x

The function  that models this  is :  y  = 10 - x

28.   Miles traveled.    Miles remaining 

                 0.                           500

               125.                         375

               350.                         150

Let the Miles Remaining be  y and the  Miles Traveled

This is similar to 26.....the function that models this is   y = 500 - x

Thanks, Melody.....here's another way that is non-trig based.....

Look at the following image :

Let's forget about the circle for a moment -  we'll return to this in a second.....

Let us place the side with length 6 along the x axis such that its endpoints lie at (-3,0)  and (3,0)...let's label these as  A and C

Let the altitude of the triangle lie along the y axis......this will actually form two right triangles.....each with a hypotenuse of 5

So....using the Pythagorean Theorem the altitude of our triangle will be sqrt [ 5^2 - 3^2]  = sqrt [ 16]  = 4

And let us call the altitude BD

Now....the center of our circle will lie on  AD....and let us call this point (0, y)

And.......the radius of our circle will pass through all three vertex points of triangle ABC.....this means that the distance from (0, y) to B    will equal the distance from  (0,y)  to C...and  B has the coordinates (0, 4)  and C has the coordinates (3,0)

So...using the distance formula,  we have that

sqrt  [ ( 0 - 0)^2  + ( y - 4)^2 ]  = sqrt [ (0-3)^2 + (y - 0)^2]     squaring both sides and simplifying, we have that

(y - 4)^2  =  9 + y^2

y^2 - 8y +  16  = 9 + y^2       subtract y^2 from both sides

-8y + 16 =  9                       add 8y to both sides, subtract 9 from both sides

7  = 8y      divide both sides by 8

7/8  = y....  so the center of our circle will lie at (0, 7/8)

And the distance from this point to D  is the radius   =  4 - 7/8 =  32/8 - 7/8 =  25 / 8   = 3.125 =  3 + 1/8  

And the equation of the circle will be  x^2 + (y - 7/8)^2  = (25/8)^2

l x - 2 l ≤ 5.6

We have two equations here

x - 2  ≤ 5.6            and                x  - 2  ≥   - 5.6

add  2 to both sides

x ≤ 7.6                                     x ≥  -3.6

So.....the solution set is

-3.6 ≤ x ≤ 7.6

And there are 11 integers between these two values

I assume you want   1 / (16/9)

Note that we can actually wirte this as   (1/1)  / (16/9)

Since we're dividing fractions......invert the fraction in the denominator and multiply by the fraction in the numerator :

(1/1) * (9/16)  =    9 / 16

Thanks, hectictar......!!!!

Another thing to realize is that circles are to one another as the squares of their radiuses.....

Thus.....the smaller circle has a radius of 1/2 the larger circle...so...its area  = (1/2)^2  = 1/4  that of the larger circle

Here's another method....

The sides are all equal, so one side  = 148/ 4  = 37

And the diagonals meet at right angles......so  1/2 the length of the known diagonal = 1/2 * 24  = 12

And this half diagonal  and one of the rhombus sides will form the leg and hypotenuse of a right triangle.....and the other half diagonal will form the other leg.....and its length  =  sqrt [ 37^2 - 12^2]  = sqrt [ 1225 ] =  35

So......the length of the other diagonal  = 2 * 35  = 70

Mmmmm....and all along I thought "Q.E.D."   referred to  "Quit Eating Dinner !!! "